Is there a neat formula for the volume of a tetrahedron on $S^3$?

On the volume of a hyperbolic and spherical tetrahedron, by Murakami and Yano. The volume is obtained as a linear combination of dilogarithms and squares of logarithms. The origin of their formula is really interesting: Asymptotics of quantum $6j$ symbols. (These asymptotics have also been studied by many other people: D. Thurston, Roberts, Woodward, Frohman, Kania-Bartoszynska, etc.)

Note that the 3-dimensional formula has to be much more complicated. The 2-dimensional formula comes from Euler characteristic and Gauss-Bonnet, but the Euler characteristic of the 3-sphere, or any odd-dimensional manifold, vanishes. In fact every characteristic class of a 3-sphere vanishes, because the tangent bundle is trivial. There can't be a purely linear treatment of volumes in isotropic spaces in odd dimensions. In even dimensions, there is always a purely linear extension from lower dimensions using generalized Gauss-Bonnet.

Nice answer, Greg. I looked at the linked paper and was sufficiently intimidated. I just want to point out, again, that for those (like me) who have a phobia of differential geometry, and hence don't want to use (generalized)Gauss-Bonnet, it is easy to see, using inclusion-exclusion, that the formula in even dimensions is a neat linear combination of the formulas in lower dimensions.