A face of the simplex is visible from just those points outside the half-space defined by that face. That’s just because the simplex is convex!

The simplex lies in the subspace S = {<s₀, s₁, ... s_n> | Σs_i = ε}. The simplex proper is for those points of S all of whose coordinates are positive.

Here is at least a Monte Carlo algorithm to compute the Aⁿ_k’s. We want to efficiently sample the distribution of directions in the subspace S taking care to make the probability of a region of directions proportional to the measure of that region. We use Gaussian distributions to overcome an efficiency problem with the more familiar schemes for choosing random directions in lower dimensions. Consider the spherical Gaussian probability distributions about the origin with large standard deviations in both Rⁿ⁺¹ and S. Choosing n+1 random normal deviates, d₀, d₁, ... d_n, and interpreting them as the coordinates of a point in Rⁿ⁺¹, gives us a sample from the n+1 dimensional Gaussian cloud. Projecting it onto the closest point in S, whose coordinates are e_i = d_i − Σd_j/(n+1) + ε/(n+1), gives us a sample from the n dimensional Gaussian cloud in S. (Soon we shall let ε go to 0.) Spherical symmetry ensures that our sample in S has the correct distribution. Notice that this amounts to subtracting the mean of the deviates from each deviate. This point e in S is in half-space k just in case e_k > 0.

Thus the (unconventional) magnitude of the angle, Aⁿ_k, defined by k faces of a regular n-simplex is the probability that the mean of n+1 samples drawn from a normal distribution will be greater than the first k of those samples.
Aⁿ_k = P(∀j(0≤j<k ⇒ Σ_id_i/(n+1) < d_j)).

This Monte Carlo program computes and reports an array, h, of counts. “n” is a compile time constant. “h[k]” is the count of the sample points that fall in k half-spaces, but no more. The program proceeds to compute Aⁿ_k from these statistics according to the ideas below.

We will need the binomial coefficients: _nC_k = n!/(k!(n−k)!) which is the number of sub-sets of size k from a set of size n. This is for n≥0 and 0≤k≤n. For k outside this range _nC_k = 0.

Let Q be the set of half-spaces in S of the form x_i > 0 for some i such that 0≤i≤n. Q has n+1 members. Let the letters p & q denote subsets of Q and let H_p denote the number of samples in ∩p. There are _n+1C_k such k-hedral angles. For each subset of k half-spaces there is a particular k-hedral angle of size Aⁿ_k. For any j<k, a sample counted in h[j] belongs to each of these _nC_k angles.

Thinking out Loud

To evaluate A⁹₄: We consider the total sample count for all quadrahedral angles and then divide by the number of these angles. Note that some samples will fall in multiple such angles and they must thus be counted as many times as there are such angles. h[3] contributes nothing for every sample that concerns us falls in at least 4 half-spaces. Samples reported by h[4] each contribute one to our total for a sample that is in just 4 half spaces falls in exactly one of our angles. Each sample from the h[5] population falls in just 5 quadrahedral angles. Thus 5h[5] = ₅C₄h[5] is included in out total. For each j from 4 to 10 we accumulate _jC₄h[j] in our total. We then divide by the number of quadrahedral angles which is ₁₀C₄.

Concisely: Aⁿ_k ≃ (Σ_jh[j](_jC_k))/(_n+1C_k Σ_jh[j]).

Program yields:

100000000 samples for the 3-simplex

j	h[j]
0	0
1	17550564
2	64899780
3	17549656
4	0

k	A3k	α3k	C
0	1.00000	0.00000	1
1	0.50000	1.00000	4
2	0.19591	1.23097	6
3	0.04387	0.55134	4
4	0.00000	0.00000	1

100000000 samples for the 4-simplex

j	h[j]
0	0
1	4894463
2	45106884
3	45104700
4	4893953
5	0

k	A4k	α4k	C
0	1.00000	0.00000	1
1	0.50000	0.99999	5
2	0.20978	1.31812	10
3	0.06468	0.81280	10
4	0.00979	0.19321	5
5	0.00000	0.00000	1

n	cos−1(1/n)	our estimate for α22
2	1.04720	1.04704
3	1.23096	1.23097
4	1.31812	1.31812

Here is the program output for 10⁸ samples in 24D.
You can see that this did not produce a value for α²⁴₂₄.