Rotate the figure so that the $z$-axis passes through two opposite vertices, $N$ and $S$. Then the edges ---three "northern" and three "southern"--- adjacent to either $N$ or $S$ lie on three uniformly-arranged planes through the $z$-axis. Let the complete great circles determined by these planes sub-divide your six "sides" into twelve congruent triangular regions. Then each region's "azimuth range" covers exactly one-sixth of a circle. That's the easy part.
The remaining six of the figure's original edges ---each separating one northern triangular region from one southern region--- form a "belt" that zig-zags above and below the equatorial $xy$-plane; let $\psi$ be the elevation angle at one of its highest points. One extreme of the "elevation range" of a region for a given azimuth is determined by this belt (the other extreme is $\pi/2$ for a northern region, and $-\pi/2$ for a southern region). That's the trickier part.
With an appropriate rotation about the $z$-axis, we can take the belt vertices to be $P_k(\;\cos\psi \; \cos\frac{k\pi}{3},\;\cos\psi \; \sin\frac{k\pi}{3},\;(-1)^k\sin\psi\;)$ for $k=0,1,2,3,4,5$. The great circle containing the belt-edge $P_k P_{k+1}$ is the intersection of the sphere with the plane through those vertices and the origin; thus, its the collection of sphere-points orthogonal to the normal to that plane, with azimuth-elevation coordinates $(\phi,\theta)$ satisfying this equation $ \begin{bmatrix}\cos\theta\cos\phi \\ \cos\theta\sin\phi \\ \sin\theta \end{bmatrix} \cdot \left( P_k \times P_{k+1} \right) = 0 $
(Tip of the hat to @joriki and his answer to your more-basic version of this question.) That is, $ \begin{eqnarray*} 0 &=&(-1)^k \cos\psi \sin\psi \left(\sin\frac{k\pi}{3}+\sin\frac{(k+1)\pi}{3}\right) \cos\theta\cos\phi \\ &-& (-1)^k \cos\psi \sin\psi \left(\cos\frac{k\pi}{3}+\cos\frac{(k+1)\pi}{3}\right) \cos\theta\sin\phi \\ &-& \cos^2\psi \left(\sin\frac{(k+1)\pi}{3}\cos\frac{k\pi}{3}-\cos\frac{(k+1)\pi}{3}\sin\frac{k\pi}{3}\right) \sin\theta \end{eqnarray*} $
This collapses to $ \tan\theta = (-1)^k \; \left( 2 \tan\psi \right) \; \sin\left(\frac{(2k+1)\pi}{6} - \phi\right) $ (Sanity check: At the endpoints $\phi_{+} = \frac{k\pi}{3}$ and $\phi_{-} = \frac{(k+1)\pi}{3}$ of a belt-edge, we get $\tan\theta_{\pm} = \pm (-1)^k \tan\psi$, so that $\theta_{\pm} = \pm(-1)^k\psi$, as desired. Also, at the midpoint $\phi_{\star} = \frac{(2k+1)\pi}{6}$, where the edge crosses the $xy$-plane, we get the correct (lack of) elevation, $\theta_{\star} = 0$.)
Now, the edge from $N$ to nearby belt-vertex (like any edge in the original figure) has arc-measure $\mathrm{acos}\frac{1}{3}$; our $\psi$ ---which measures from equator to belt-vertex--- is the complement of this angle, so that $\tan\psi = \frac{1}{2\sqrt{2}}$ and the final formula becomes $ \tan\theta = \frac{(-1)^k}{\sqrt{2}} \sin\left(\frac{(2k+1)\pi}{6} - \phi\right) $
This gives the non-obvious extreme of the elevation range for points in one of the triangular regions. Since these regions merely sub-divide your six "sides", you have a parameterization of those sides. (I'll leave it to you to transform the parameterization if you absolutely require the $z$-axis to pass through opposing sides rather than opposing vertices.)