Hi!

But maybe you need the solution to another, similar problem that consists in finding the solid angle of a pixel lying in the z=0 plane and orthogonally projected onto the hemisphere, but couldn't find a page with that computation?

The configuration for a cube map is that pixels are lying on a plane z=1 such as $p'=(x,y,1), -1<x<1, -1<y<1$ and we project back onto the unit hemisphere by "normalizing" the vector $p=\frac{\mathbf{p'}}{|\mathbf{p'}|}=\frac{(x,y,1)}{\sqrt{1+x^2+y^2}}$ as shown on the figure below:
The idea is to compute the area of a small element of surface on the hemisphere as we make it vary on the plane, we do that by computing the partial derivatives of $\mathbf{p}$ along x and y that give us the vectors $\frac{\partial \mathbf{p}}{\partial x}$ and $\frac{\partial \mathbf{p}}{\partial y}$.