$n$ points are placed on an $n-2$-sphere so that the smallest angle from the centre between any pair of the points is maximised. What is this smallest angle?
$n=1 \ \ \ \cos^{-1}{1}\\ n=2 \ \ \ \cos^{-1}{-1}\\ n=3 \ \ \ \cos^{-1}{\frac{-1}2}\\ n=4 \ \ \ \cos^{-1}{\frac{-1}3}$
The smallest angle will be the same for any $n$ points on a $p$-sphere where $p> n-2$ as when $p=n-2$.
On
I guess that the $n$ points on the $(n-2)$-sphere $S^{n-2}\subset {\mathbb R}^{n-1}$ would have to form a regular $(n-1)$-simplex $\Sigma_{n-1}\subset S^{n-2}$ (one would have to prove this). The regular simplex is one of the three Platonic solids that exist in all dimensions (the other two are the cube and the octahedron).
The regular simplices in higher dimensions can be recursively constructed in the following way: When $s_1$, $s_2$, $\ldots$, $s_n\in S^{n-2}$ are the vertices of $\Sigma_{n-1}$ then the $n+1$ vertices $x_i$ of $\Sigma_n \subset {\mathbb R}^n= {\mathbb R}^{n-1}\times{\mathbb R}$ have the form $$x_i=-\lambda e_{n+1}+\mu s_i\quad(1\leq i\leq n)\ ,\qquad x_{n+1}=e_{n+1}\ ,$$ where $e_{n+1}$ is the new basis vector. Taking care of $|x_i|=1$, $x_i\cdot x_{n+1}=x_i\cdot x_j$ $ \ (1\leq i < j\leq n)$ will determine $\lambda$ and $\mu$ and will also lead to a recursion for $\cos\phi_n$.
I don't understand why you say "there is no 4D equivalent to a platonic solid with $5$ vertices". The angles you give are the angles in the regular $(n-1)$-simplices. The vertices of a regular $(n-1)$-simplex lie on an $(n-2)$-sphere, and it seems clear that the corresponding angles are maximal. Two vertices and the centre of a regular $(n-1)$-simplex can be placed at $(n,0,0,\dotsc)$, $(0,n,0,\dotsc)$ and $(1,1,1,\dotsc)$, respectively; the vectors from the centre to the vertices are $(n-1,-1,-1,\dotsc)$ and $(-1,n-1,-1,\dotsc)$, with cosine $-n/((n-1)^2+n-1)$ $=-1/(n-1)$. So the sequence continues as expected, $\arccos(-1/(n-1))$.