Upon some reflection, the answer is that no, they most definitely do not.

A first way to see this is to notice that a hypersphere has a very "regular" structure. Suppose there were some $2(d-1)$ basis (normalised) vectors $\{\mathbf v_1,...,\mathbf v_{2d-2}\}\subset\mathbb R^{d^2-1}$ forming a sphere containing the pure states. This obviously cannot be the case, because we know that there are $d^2-1$ different orthogonal directions which contain pure states.

Indeed, given any orthogonal basis of Hermitian traceless operators $\{\boldsymbol\sigma_j\}_{j=1}^{d^2-1}$, and any normal versor $\hat{\mathbf n}\in\mathbb R^{d^2-1}$ with $\|\hat{\mathbf n}\|=1$, there are always pure states in the direction $\boldsymbol\sigma_{\hat{\mathbf n}}\equiv\hat{\mathbf n}\cdot\boldsymbol\sigma$ (this follows from the fact that the operator norm satisfies $\|\boldsymbol\sigma_{\hat{\mathbf n}}\|>0$ and the characterisation in spherical coordinates explained e.g. here). It follows that there cannot be less than $d^2-1$ elements of $\mathbb R^{d^2-1}$ whose span contains the set of pure states.

A concrete example: three-level systems

Consider a generic pure state of a three level system: $$|\psi\rangle=\cos\alpha|0\rangle+e^{i\phi}\sin\alpha \cos\beta |1\rangle+e^{i\theta}\sin\alpha\sin\beta|2\rangle,$$ for all $\alpha,\beta,\phi,\theta\in\mathbb R$. Let us also use the standard operatorial basis for this space (the matrices used at the bottom of this answer):

$$ Z^{(1)}=\sqrt{\frac{3}{2}}\begin{pmatrix}1 & 0&0 \\ 0 & -1&0\\0&0&0\end{pmatrix}, \quad Z^{(2)}=\sqrt{\frac{3}{6}}\begin{pmatrix}1 & 0&0 \\ 0 & 1&0\\0&0&-2\end{pmatrix}, $$ \begin{align} X^{(12)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 1&0 \\ 1 & 0&0\\0&0&0\end{pmatrix}, \quad X^{(13)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&1 \\ 0 & 0&0\\1&0&0\end{pmatrix}, \quad X^{(23)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&0 \\ 0 & 0&1\\0&1&0\end{pmatrix} \end{align} \begin{align} Y^{(12)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & -i&0 \\ i & 0&0\\0&0&0\end{pmatrix}, \quad Y^{(13)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&-i \\ 0 & 0&0\\i&0&0\end{pmatrix}, \quad Y^{(23)}=\sqrt{\frac{3}{2}}\begin{pmatrix}0 & 0&0 \\ 0 & 0&-i\\0&i&0\end{pmatrix}. \end{align} Then, the surface covered by the pure states in $\mathbb R^8$ has the following parametrisation: \begin{cases} \langle Z^{(1)}\rangle&=\sqrt{3/2} (\cos^2\alpha-\sin^2\alpha\cos^2\beta),\\ \langle Z^{(2)}\rangle&=\sqrt{3/6} [\cos^2\alpha+\sin^2\alpha(\cos^2\beta-2\sin^2\beta)],\\\hline \langle X^{(12)}\rangle&=\sqrt{3/2}\sin(2\alpha)\cos\beta \cos\phi,\\ \langle X^{(13)}\rangle&=\sqrt{3/2}\sin(2\alpha)\sin\beta \cos\theta,\\ \langle X^{(23)}\rangle&=\sqrt{3/2}\sin^2(\alpha)\sin(2\beta) \cos(\phi-\theta),\\\hline \langle Y^{(12)}\rangle&=\sqrt{3/2}\sin(2\alpha)\cos\beta \sin\phi,\\ \langle Y^{(13)}\rangle&=\sqrt{3/2}\sin(2\alpha)\sin\beta \sin\theta,\\ \langle Y^{(23)}\rangle&=\sqrt{3/2}\sin^2(\alpha)\sin(2\beta) \sin(\phi-\theta).\tag A \end{cases}

To easily check that these points do indeed lie on a hypersphere (which we also know from this answer must have a radius of $\sqrt2$), just run the following snippet in Mathematica:

$Assumptions = Element[{\[Alpha], \[Beta], \[Theta], \[Phi]}, Reals];

expvalZ1[\[Alpha]_, \[Beta]_, \[Phi]_:0, \[Theta]_:0] = Sqrt[3/2]*(Cos[\[Alpha]]^2 - Sin[\[Alpha]]^2*Cos[\[Beta]]^2); 
expvalZ2[\[Alpha]_, \[Beta]_, \[Phi]_:0, \[Theta]_:0] = Sqrt[3/6]*(Cos[\[Alpha]]^2 + Sin[\[Alpha]]^2*(Cos[\[Beta]]^2 - 2*Sin[\[Beta]]^2)); 
expvalX12[\[Alpha]_, \[Beta]_, \[Phi]_, \[Theta]_:0] = Sqrt[3/2]*Sin[2*\[Alpha]]*Cos[\[Beta]]*Cos[\[Phi]]; 
expvalX13[\[Alpha]_, \[Beta]_, \[Phi]_:0, \[Theta]_] = Sqrt[3/2]*Sin[2*\[Alpha]]*Sin[\[Beta]]*Cos[\[Theta]]; 
expvalX23[\[Alpha]_, \[Beta]_, \[Phi]_, \[Theta]_] = Sqrt[3/2]*Sin[\[Alpha]]^2*Sin[2*\[Beta]]*Cos[\[Phi] - \[Theta]]; 
expvalY12[\[Alpha]_, \[Beta]_, \[Phi]_, \[Theta]_:0] = Sqrt[3/2]*Sin[2*\[Alpha]]*Cos[\[Beta]]*Sin[\[Phi]]; 
expvalY13[\[Alpha]_, \[Beta]_, \[Phi]_:0, \[Theta]_] = Sqrt[3/2]*Sin[2*\[Alpha]]*Sin[\[Beta]]*Sin[\[Theta]]; 
expvalY23[\[Alpha]_, \[Beta]_, \[Phi]_, \[Theta]_] = Sqrt[3/2]*Sin[\[Alpha]]^2*Sin[2*\[Beta]]*Sin[\[Phi] - \[Theta]];

Simplify[Total[(#1[\[Alpha], \[Beta], \[Phi], \[Theta]]^2 & ) /@ {expvalZ1, expvalZ2, expvalX12, expvalX13, expvalX23, expvalY12, expvalY13, expvalY23}]]

Now, what sort of $4$-dimensional surface in $\mathbb R^8$ is this? I don't know a full answer to this, but if it was a hypersphere, then the sections would look like lower-dimensional hyperspheres (e.g. plotting only in two dimensions should result in a series of circles). This is most definitely not the case.

Indeed, for completeness, here is what some of the sections look like (I will again use Mathematica for the plotting):

$Z^{(1)}, Z^{(2)}$ section

These coordinates only depend on the $\alpha,\beta$ parameters. It follows that the full space is contained in a sort of "tubular region" with the following triangular 2D section:

ParametricPlot[{expvalZ1[\[Alpha], \[Beta]], 
  expvalZ2[\[Alpha], \[Beta]]}, {\[Alpha], 0, Pi}, {\[Beta], 0, Pi}]

So again, most definitely not what a hypersphere would look like.

Two-dimensional $X^{(ij)}$ sections

The two-dimensional sections obtained using two of the three available $X^{(ij)}$ coordinates look like ellipses when varying $\alpha,\beta$ for fixed $\theta,\phi$ (with the principal axes varying with $\theta,\phi$). Interestingly, if we instead vary $\theta,\phi$ for fixed $\alpha,\beta$, the sections look instead like rectangles. I won't include the plots, but you can use the following code to show these sections:

Manipulate[
 ParametricPlot[{expvalX12[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalX13[\[Alpha], \[Beta], \[Theta], \[Phi]]}, {\[Alpha], 0, 
   Pi}, {\[Beta], 0, Pi}, PerformanceGoal -> "Quality", 
  PlotRange -> {{-2, 2}, {-2, 2}}],
 {{\[Theta], 0}, 0, Pi, 0.01, Appearance -> "Labeled"}, {{\[Phi], 0}, 
  0, Pi, 0.01, Appearance -> "Labeled"}]

Manipulate[
 ParametricPlot[{expvalX12[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalX13[\[Alpha], \[Beta], \[Theta], \[Phi]]}, {\[Theta], 0, 
   Pi}, {\[Phi], 0, Pi}, PerformanceGoal -> "Quality", 
  PlotRange -> {{-2, 2}, {-2, 2}}],
 {{\[Alpha], 0}, 0, Pi, 0.01, Appearance -> "Labeled"}, {{\[Beta], 0},
   0, Pi, 0.01, Appearance -> "Labeled"}]

Three-dimensional $X^{(ij)}$ section

This is where it starts to look pretty cool.

If we vary $\alpha,\beta$ for $\theta,\phi$ fixed, we get this nice 3D shape (sorry for the poor quality, SE doesn't allow more than 2MB for images):

Manipulate[
 ParametricPlot3D[{expvalX12[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalX13[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalX23[\[Alpha], \[Beta], \[Theta], \[Phi]]}, {\[Alpha], 0, 
   Pi}, {\[Beta], 0, Pi}, PerformanceGoal -> "Quality", 
  PlotRange -> Evaluate[{{-#, #}, {-#, #}, {-#, #}} &@Sqrt@2]],
 {{\[Theta], 0}, 0, Pi, 0.01, Appearance -> "Labeled"}, {{\[Phi], 0}, 
  0, Pi, 0.01, Appearance -> "Labeled"}, ControlPlacement -> Right]

Changing the values of $\phi,\theta$ mostly just changes the scales of the surface, without modifying its structure. Here is a better quality, still picture of this surface:

The $Y^{(ij)}$ sections have pretty much the same structure, modulo changing the values of $\phi,\theta$.

So yea, in conclusion, definitely not a hypersphere.

PS: For anyone that wants to see how the pure states move in the overall $8$-dimensional surface, you can use the following code to see the point corresponding to the given state in the $X,Y$ and $Z$ sections at the same time, changing the values of the $\alpha,\beta,\theta,\phi$ parameters (the blue dot represents the state for the given values of the parameters):

pointOnXSection[\[Alpha]_, \[Beta]_, \[Theta]_, \[Phi]_] := \
{expvalX12[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalX13[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalX23[\[Alpha], \[Beta], \[Theta], \[Phi]]};
pointOnYSection[\[Alpha]_, \[Beta]_, \[Theta]_, \[Phi]_] := \
{expvalY12[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalY13[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalY23[\[Alpha], \[Beta], \[Theta], \[Phi]]};
pointOnZSection[\[Alpha]_, \[Beta]_, \[Theta]_, \[Phi]_] := \
{expvalZ1[\[Alpha], \[Beta], \[Theta], \[Phi]], 
   expvalZ2[\[Alpha], \[Beta], \[Theta], \[Phi]]};
Manipulate[
 GraphicsRow[{
   Show[
    ParametricPlot3D[
     pointOnXSection[\[Alpha], \[Beta], \[Theta], \[Phi]], {\[Alpha], 
      0, Pi}, {\[Beta], 0, Pi}, PerformanceGoal -> "Quality", 
     PlotRange -> Evaluate[{{-#, #}, {-#, #}, {-#, #}} &@Sqrt@2], 
     AxesLabel -> {1, 2, 3}, ImageSize -> 400, 
     PlotStyle -> Directive[[email protected]], RotationAction -> "Clip"],
    Graphics3D[{Blue, [email protected], 
      Point@pointOnXSection[dot\[Alpha], 
        dot\[Beta], \[Theta], \[Phi]]}]
    ],
   Show[
    ParametricPlot3D[
     pointOnYSection[\[Alpha], \[Beta], \[Theta], \[Phi]], {\[Alpha], 
      0, Pi}, {\[Beta], 0, Pi}, PerformanceGoal -> "Quality", 
     PlotRange -> Evaluate[{{-#, #}, {-#, #}, {-#, #}} &@Sqrt@2], 
     AxesLabel -> {1, 2, 3}, ImageSize -> 400, 
     PlotStyle -> Directive[[email protected]], RotationAction -> "Clip"],
    Graphics3D[{Blue, [email protected], 
      Point@pointOnYSection[dot\[Alpha], 
        dot\[Beta], \[Theta], \[Phi]]}]
    ],
   Show[
    ParametricPlot[
     pointOnZSection[\[Alpha], \[Beta], \[Theta], \[Phi]], {\[Alpha], 
      0, Pi}, {\[Beta], 0, Pi}, PerformanceGoal -> "Quality", 
     PlotRange -> Evaluate[{{-#, #}, {-#, #}} &@Sqrt@2], 
     AxesLabel -> {1, 2, 3}, ImageSize -> 300, 
     PlotStyle -> Directive[[email protected]]],
    Graphics[{Blue, [email protected], 
      Point@pointOnZSection[dot\[Alpha], 
        dot\[Beta], \[Theta], \[Phi]]}]
    ]
   }, ImageSize -> 800],
 {{\[Theta], 0}, 0, Pi, 0.01, Appearance -> "Labeled"}, {{\[Phi], 0}, 
  0, Pi, 0.01, Appearance -> "Labeled"},
 {{dot\[Alpha], 0, "\[Alpha]"}, 0, Pi, 0.01, 
  Appearance -> "Labeled"}, {{dot\[Beta], 0, "\[Beta]"}, 0, Pi, 0.01, 
  Appearance -> "Labeled"},
 ControlPlacement -> Right]