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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. In other words, a lower-dimensional hypersphere embedded in the full space would mean that the set of pure states is a subset of the span of less than $d^2-1$ orthogonal vectors. But we know that we need all $d^2-1$ orthogonal directions to describe generic pure states (see also paragraph below), so this cannot be true.

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.

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):

The easiest way to see this is to observe that there are $d^2-1$ orthogonal directions in the Bloch representation (i.e. orthogonal Hermitian traceless operators) containing pure states. This means that that the pure states are not contained in any linear subspace of dimension less than $d^2-1$, and thus in particular cannot be contained in any lower-dimensional hypersphere.

More specifically, given any orthogonal basis of Hermitian traceless operators $\{\boldsymbol\sigma_j\}_{j=1}^{d^2-1}$, and any 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.

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):

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.

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. In other words, a lower-dimensional hypersphere embedded in the full space would mean that the set of pure states is a subset of the span of less than $d^2-1$ orthogonal vectors. But we know that we need all $d^2-1$ orthogonal directions to describe generic pure states (see also paragraph below), so this cannot be true.

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]