Extending the relationship between $SU(2)$ and $SO(3)$ to higher dimensions
The analogy for understanding single-qubit gates in terms of $SO(3)$ is provided by an accidental isomorphism $Spin(3)\cong SU(2)$ where the spin group $Spin(n)$ is the double cover of $SO(n)$. As the name suggests, the isomorphism is not part of a recurring pattern, so the answer to the first question is negative in general.
However, $Spin(3)\cong SU(2)$ isn't the only accidental isomorphism. Other interesting examples include $Spin(4) \cong SU(2) \times SU(2)$ which is related to the fact any 4D real rotation can be described by two quaternions (one acting by left-multiplication and the other by right-multiplication) and $Spin(6) \cong SU(4)$ (see this answer for details of this isomorphism). Nevertheless, real rotations in 4D and 6D are not as intuitive as those in 2D and 3D. A better approach to gaining intuitive understanding of multi-qubit gates is implicit in the second question: identify an interesting type of gate which is easier to describe.
General remark about understanding unitary gates as rotations
A key realization that aids in understanding quantum gates in terms of rotations is that all operators in $SU(n)$ can be diagonalized in $\mathbb{C}^n$, but a generic element of $SO(n)$ cannot be diagonalized in $\mathbb{R}^n$. A geometric consequence of this fact is that a generic element of $SO(n)$ changes direction of some basis vectors. By contrast, in the appropriate basis the action of a unitary operator is the multiplication of all vector components by various scalar phase factors. In a sense, the action of the operator is decomposed into 2D rotations in the field of complex numbers.
Origin of the $\pi/4$ angle in the formula for SWAP
Perhaps the most familiar example of a unitary gate understood via its counterpart in $SO(3)$ is the single-qubit rotation
$$
R_{\hat n}(\alpha) = \exp\left(-i\frac{\alpha}{2}(n_x X + n_y Y + n_z Z)\right) = I\cos\frac{\alpha}{2} -i (n_x X + n_y Y + n_z Z)\sin\frac{\alpha}{2}
$$
where $\hat n = (n_x, n_y, n_z)$ is a real 3-vector of unit length. This can be generalized as
$$
\exp\left(i\frac{\beta}{2} A\right) = I \cos\frac{\beta}{2} + i A \sin\frac{\beta}{2}\tag1
$$
where $\beta\in\mathbb{R}$ and $A$ is a matrix such that $A^2 = I$ (see exercise 4.2. on p.175 in section 4.2 of Nielsen & Chuang). This formula has the benefit of making it clear which values of the angle $\beta$ correspond to the identity and to the unitary $A$.
Now, before we write the SWAP gate in the form of $(1)$ let us first generalize it to a single-parameter group
$$
\text{SWAP}(\theta) = \exp\left(i \frac{\theta}{4} (XX + YY + ZZ)\right).
$$
Unfortunately, the naive attempt of writing the gate as $(1)$ by setting $A = XX + YY + ZZ$ fails because as we see from the multiplication table
$$
\begin{array}{c|ccc}
& XX & YY & ZZ\\
\hline
XX & II & -ZZ & -YY\\
YY & -ZZ & II & -XX\\
ZZ & -YY & -XX & II
\end{array}
$$
the non-identity terms in $(XX + YY + ZZ)^2$ do not cancel. However, since the global phase has no physical meaning, we can multiply any unitary matrix by a phase factor without changing the corresponding quantum gate. In our case, we multiply by $\exp\left(i\frac{\theta}{4}\right)$ to obtain
$$
\text{SWAP}(\theta) = \exp\left(i \frac{\theta}{4} (II + XX + YY + ZZ)\right)
$$
with exponent that squares to a multiple of identity as can be seen from the new multiplication table
$$
\begin{array}{c|cccc}
& II & XX & YY & ZZ\\
\hline
II & II & XX & YY & ZZ\\
XX & XX & II & -ZZ & -YY\\
YY & YY & -ZZ & II & -XX\\
ZZ & ZZ & -YY & -XX & II
\end{array}
$$
The table shows that $(II + XX + YY + ZZ)^2 = 4II$ and so
$$
\text{SWAP}(\theta) = II \cos\frac{\theta}{2} + i \frac{II + XX + YY + ZZ}{2} \sin\frac{\theta}{2}.
$$
This explains why $\pi$ is divided by $4$ in the exponential formula given in the question. One factor of $2$ normalizes $II + XX + YY + ZZ$ and the other factor of $2$ comes from formula $(1)$. Consequently, $\text{SWAP}(\theta)$ behaves analogously to $R_{\hat n}(\alpha)$: rotation by $2\pi$ yields $-I$ and a rotation by $4\pi$ returns to $I$, just as expected from the "belt trick". Note however that the full SWAP takes place for $\theta=\pi$ and $\theta=3\pi$, not $2\pi$. This is analogous to how the full bit-flip is effected by $R_X(\alpha)$ for $\alpha=\pi$ and $3\pi$.
Limited application of the Bloch sphere to multi-qubit gates
Another way of understanding gates in terms of rotations is available when the gate in question acts non-trivially on a subspace of dimension two. In this case, we can use the Bloch sphere to get a limited intuitive picture of the action of the gate. For example, $\text{SWAP}(\theta)$ can be written as
$$
\text{SWAP}(\theta) = \begin{pmatrix}
e^{i\frac{\theta}{2}} & & & \\
& \cos\frac{\theta}{2} & i\sin\frac{\theta}{2} & \\
& i\sin\frac{\theta}{2} & \cos\frac{\theta}{2} & \\
& & & e^{i\frac{\theta}{2}} \\
\end{pmatrix}
$$
and we recognize the middle $2\times 2$ block as $R_X(-\theta) \in SU(2)$. In other words, if we relabel the North Pole of the Bloch sphere to $|10\rangle$ and the South Pole to $|01\rangle$ then the action of $\text{SWAP}(\theta)$ on the $\mathrm{span}(|01\rangle, |10\rangle)$ subspace is the same as the action of $R_X(-\theta)$ on a qubit. Note however, that this interpretation ignores the change in relative phase imparted by the gate between $\mathrm{span}(|01\rangle, |10\rangle)$ and $\mathrm{span}(|00\rangle, |11\rangle)$.
