My answer to the "Favorite Equations" question was the Pythagorean Theorem for Right-Corner Tetrahedra:
Euclidean: $A^2+B^2+C^2=D^2$
Hyperbolic: $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}−\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\cos\frac{D}{2}$
Spherical: $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\cos\frac{D}{2}$
where A, B, C are the areas of the "leg-faces" and D is the area of the "hypotenuse-face".
For right-corner simplices in higher Euclidean dimensions, we have that the sum of the squares of the content of leg-simplices equals the square of the content of the hypotenuse-simplex.
I mentioned not knowing the non-Euclidean counterparts of the generalization. After a couple of days of toying, I find these counterparts elusive. (It would probably help if I were better-versed in differential geometry.) So, I turn to MO to ask:
What are the non-Euclidean analogues in higher dimensions?
(I'm particularly interested in the case relating volumes in a hyperbolic 4-simplex.)
While the Pythagorean Theorems for Euclidean Simplices progress in a straightforward manner (just add another leg-content-square), the Pythagorean Theorem for Hyperbolic Tetrahedra already diverges somewhat dramatically from its 2-dimensional counterpart, $\cosh a \cosh b = \cosh c$. So, it's not even clear to me what form the relations would take in general.
Edit to add a couple of references
"The Laws of Cosines for Non-Euclidean Tetrahedra" (PDF) (by me) derives a Law of Cosines for which each non-Euclidean Pythagorean Theorem is a special case. I do not know if these results exist elsewhere in the literature. (BTW: There's a lot of unnecessary equation manipulation shown; I was using this as an opportunity to learn LaTeX. :)
A passage in above jumps into a discussion of "pseudofaces" by making passing reference to the Euclidean case. You can read about Euclidean pseudofaces and how I find them useful here:
"Heron-like Results for [Euclidean] Tetrahedral Volume" (by me).
Edit2: A Special Case (Edit 3: A strikingly-similar pair of volume formulas.)
Consider the special case of a right-corner hyperbolic $4$-simplex whose "legs" are congruent right-corner tetrahedra with isosceles right-triangle faces; the "hypotenuse" is a regular tetrahedron with equilateral faces. Each face of the simplex's hypotenuse is an hypotenuse-face of one of the simplex's legs.
Let the volume of each leg-tetrahedron be $L$ and the volume of the hypotenuse-tetrahedron be $H$. Then we have these formulas:
$$L=3 \int_{\rm{acos}\sqrt{x}}^{\rm{acos}\sqrt{\frac{1}{3}}} \rm{atanh}\sqrt{\frac{3\cos^2t-1}{1-\cos^2t}} \, \mathrm{d}t$$
$$H=6\int_{\rm{acos}x}^{\rm{acos}\frac{1}{3}} \rm{atanh}\sqrt{\frac{3\cos t-1}{1-\cos t}} \, \mathrm{d}t$$
where $\frac{1}{3} \le x = \cos^2{\theta_L} = \cos{\theta_H} \le \frac{1}{2}$, for $\theta_L$ the acute dihedral angle in the leg-tetrahedron and $\theta_H$ the dihedral angle in the hypotenuse-tetrahedron.
The similarity in form is rather intriguing, though not necessarily encouraging: swapping "$\cos^2 t$" for "$\cos t$" in an integrand, or "$\sqrt{x}$" for "$x$" in a limit, can completely change the nature of an integral, so there's no reason to expect a straightforward relationship between these formulas. And yet, Pythagoras beckons: there must be some connection here!
Now, it's possible to write series for $H$ and $L$, then invert the second series, then substitute back in to the first series to arrive at a series for $H$ in terms of $L$.
$$H = \frac{4}{2!\;3!} M^3 + \frac{18}{3!\;5!}M^5 - \frac{ 918 }{4!\;7!}M^7 + \frac{24786}{5!\;9!}M^9 - \frac{ 6018759 }{8 \cdot 6!\;11!} M^{11} - \frac{ 8233607961 }{80\cdot 7!\;13!} M^{13} - \cdots $$ where $M := (6L)^{1/3}$. This isn't the relation I'm seeking, but observe that, for infinitesimal $L$, we have $H \approx 2 L$; that is, $H^2 \approx 4L^2 = L^2 + L^2 + L^2 + L^2$, which is the corresponding Pythagorean relation for Euclidean $4$-simplices. Even so, despite my best efforts of playing with these series, I have yet to get any insights into the nature of a non-infinitesimal connection.
I'll close here by mentioning a special-special case: at the extreme, a quadrupally-asymptotic right-corner $4$-simplex has legs that are triply-asymptotic right-corner tetrahedra with doubly-asymptotic right-triangle leg-faces; the simplex's hypotenuse is a quadrupally-asymptotic regular tetrahedron with triply-asymptotic equilateral faces. The volumes of the components (with parameter $x = 1/2$) attain significant values that happen to have interesting series representations of their own:
$$L^\star := \frac{1}{2} \; \Im\left(Li_2\left(\exp\frac{i\pi}{2}\right)\right) = \frac{1}{2} \; \sum_{k=1}^{\infty} \frac{1}{k^2}\sin{\frac{\pi k}{2}} = 0.45798\dots$$
$$H^\star := \Im\left(Li_2\left(\exp\frac{i\pi}{3}\right)\right) = \; \sum_{k=1}^{\infty} \frac{1}{k^2}\sin{\frac{\pi k}{3}} = 1.01494\dots$$
... where $Li_2$ is the dilogarithm.
Here, $L^\star$ is half of Catalan's Constant, and $H^\star$ is also known in the literature. (It is, for instance, the maximum of the Clausen function $\mathrm{Cl}_2$.)
If there's going to be a Pythagorean Theorem for Hyperbolic Simplices, then it must apply to this case, ideally relating these values in the non-Euclidean Pythagorean tradition:
$$\text{function}(H^\star) = \text{combination of related functions}(L_1=L^\star;L_2=L^\star;L_3=L^\star;L_4=L^\star)$$
where the right-hand side is symmetric in the four parameters $L_i$ (representing the volumes of the four legs of the simplex), which are all set equal to $L^\star$. (When the formula is populated with infinitesimal quantities, it should collapse to the Euclidean sum-of-squares relation.) However, while the dilogarithm has many interesting properties, the connection between $H^\star$ and $L^\star$ is not obvious (to me). In a separate MO post, I note a hypergeometric series "similar" to a series for $H^\star$ (what I call "$T(1/2)$" there) that has a direct relation to Catalan's constant (and therefore $L^\star$), but this hasn't yet provided appropriate insights into relating $H^\star$ to $L^\star$ directly.
Have I perhaps lost the forest amid a bunch of trees?