I do not think that anyone knows sharp bounds, and probably no one cares.

On the other hand, there is another version of strainers where the sharp bound is obvious. The condition is $$\tilde\measuredangle(p\,{}^{a_i}_{a_j})>\tfrac\pi2+\delta$$ for a point array $a_0,\dots,a_n$. In this case the distance map $M\to \mathbb{R}^n$ defined by $$x\mapsto (|a_1-x|,\dots,|a_n-x|)$$ is bi-Lipschitz in a neighborhood of $p$ (note that we did not use $a_0$).

This is not true if $\delta\le 0$, even for $n$-dimensional Euclidean space.

One has to work bit more to introduce this type of strainers, but they provide extra flexibility. In particular they are essential in the proof stability theorem.

Your question is almost equivalent to following problem:

Find the minimal value $\delta_n$ such that $\mathbb{S}^{n-1}$ contains $2\cdot(n+1)$ points on the distance at least $\tfrac\pi2-\delta_n$ from each other.

I guess it is not hard to solve, if the oprimal configuration is centrally symmetric, then $\delta_n$ is the bound you are looking for.

On the other hand, there is another version of strainers where the sharp bound is obvious. The condition is $$\tilde\measuredangle(p\,{}^{a_i}_{a_j})>\tfrac\pi2+\delta$$ for a point array $a_0,\dots,a_n$. In this case the distance map $M\to \mathbb{R}^n$ defined by $$x\mapsto (|a_1-x|,\dots,|a_n-x|)$$ is bi-Lipschitz in a neighborhood of $p$ (note that we did not use $a_0$).

This is not true if $\delta\le 0$, even for $n$-dimensional Euclidean space.

One has to work bit more to introduce this type of strainers, but they provide extra flexibility. In particular they are essential in the proof stability theorem.