Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$.
As originally conjectured and now proved, for $d=2$ we have:
$$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$
For $d=3$, which I’ve checked up to $m=15$, the formula appears to be:
$$E(A_3^{2m})=\frac{6\ (m!)^3\ (6 m+1)!}{(2 m)!\ (3 m)!\ (4 m+3)!}$$
The first 10 values are:
$$\frac{1}{2},\frac{13}{30},\frac{323}{700},\frac{23}{42},\frac{4495}{6468},\frac{33263}{36036},\frac{65231}{51480},\frac{21460999}{12033450},\frac{30741431}{12009140},\frac{8965109}{2401828}$$
For $d=4$, which I’ve checked up to $m=12$, there appears to be a simple relationship with the $d=2$ case:
$$ E(A_4^{2m})=\frac{8\ (3(m+1))!}{16^{m+1}\ ((m+1)!)^3\ (2 m+3)} =\frac{8\ E(A_2^{2(m+1)})}{2m+3} $$
The first 9 values are:
$${\frac{9}{16},\frac{15}{32},\frac{1925}{4096},\frac{17199}{32768},\frac{82467}{131072},\frac{415701}{524288},\frac{278397405}{268435456},\frac{2998334625}{2147483648}},\frac{66083295135}{34359738368}$$
For $d=6$, which I’ve checked up to $m=6$, there seems to be a similar relationship with the $d=4$ moments:
$$E(A_6^{2m}) = \frac{16\ E(A_4^{2(m+1)})}{3m+9}$$
The first 6 values are:
$$\frac{5}{8},\frac{385}{768},\frac{1911}{4096},\frac{3927}{8192},\frac{ 138567}{262144},\frac{10311015}{16777216}$$
Something very similar holds for the even moments for the distance $s$ between two random points on the unit sphere. As discussed in the blog posts linked to in the question, I proved a general formula for the moments of $s_d$:
$$E(s_d^n) = \frac{2^{d+n-2}\ \Gamma \left(\frac{d}{2}\right) \Gamma \left(\frac{1}{2} (d+n-1)\right)}{\sqrt{\pi }\ \Gamma \left(d+\frac{n}{2}-1\right)}$$ For even moments $n=2m$ and dimensions $d=2,4$, this specialises to the central binomial coefficients and the Catalan numbers:
$$E(s_2^{2m})=\binom{2m}{m}$$ $$E(s_4^{2m})=\frac{1}{m+2}\binom{2(m+1)}{m+1} = \frac{E(s_2^{2(m+1)})}{m+2}$$
And in general, we have the relationship:
$$E(s_d^{2m}) = \frac{(d-2)}{2\ (m+d-2)}E(s_{d-2}^{2(m+1)})$$