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:

$$E(A_3^{2m})=\frac{6\ (m!)^3\ (6 m+1)!}{(2 m)!\ (3 m)!\  (4 m+3)!}$$

For $d=4$, which I’ve checked up to $m=12$, there appears to be a simple relationship with the $d=2$ case:

$$
E(A_4^{2m})=\frac{8\ (3(m+1))!}{16^{m+1}\ ((m+1)!)^3\ (2 m+3)}
=\frac{8\ E(A_2^{2(m+1)})}{2m+3}
$$

Further computations up to $d=13$ are all consistent with the following recursion formula, for $d\ge 4$:

$$
E(A_d^{2m})=\frac{4\ (d-2)^2}{(d-3)\ (4m+3d-6)} E(A_{d-2}^{2(m+1)})
$$

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:

$$E(s_d^{2m}) = \frac{(d-2)}{2\ (m+d-2)}E(s_{d-2}^{2(m+1)})$$