For $p, x_i\in \mathbb{R}^n$, with $1\le i\le n$, define $$\theta_i=\angle(x_i,\text{span}(\{x_j \;|\; j\ne i\}))$$$$\theta_i=\angle(x_i,\text{span}(\{x_j \;|\; j\ne i\})),\quad \theta=\min(\{\theta_i\})$$
$$\beta_i=\frac{\pi}{2}-\angle(p,x_i)$$$$\beta_i=\frac{\pi}{2}-\angle(p,x_i),\quad\beta=\max(\{\beta_i\})$$
CLAIM. If $\theta_i\ge \theta>0$ for every $i$, then there is at least one $i$ such that $\displaystyle \beta_i\ge \frac{\theta}{n}$. $$\sin(\beta)\ge\frac{sin(\theta)}{\sqrt{n(1+(n-1)\cos(\theta))}}$$
PROOF. Trivial if $\theta=0$, so assume $\theta>0$. Write $p$ and the $x_i$'s as $1\times n$ matrices and rescale them so that $pp^T=x_i x_i^T=1$. Let $X$ be the matrix with rows $x_1,\dots x_n$. Then $\theta>0$ implies that $X$ beingis invertible implies, therefore
$$(pX^T)(XX^T)^{-1}(pX^T)^T=pp^T=1$$
But $pX^T=(\sin(\beta_1),\dots \sin(\beta_n))$ and the inverse correlation matrix $(XX^T)^{-1}=(c_{ij})$ satisfies
$$\displaystyle c_{ij}=\frac{1}{\sin(\theta_i)\sin(\theta_j)}\cdot \frac{x'_i {x'_j}^T}{|x'_i||x'_j|}$$
where $x'_i$ is the component of $x_i$ orthogonal to $\text{span}(\{x_j \;|\; j\ne i\})$. (Different proofs can be found here and here, but a standard linear algebra text reference would be welcome.) ThisMoreover $x'_j \perp x_i$ for $j\ne i$ also implies $\text{span}(\{x'_j \;|\; j\ne i\}$ is the hyperplane orthogonal to $x_i$, and thus $\angle(x'_i,\text{span}(\{x'_j\;|\; j\ne i\}))=\angle(\text{span}(\{x_j \;|\; j\ne i\}),x_i)=\theta_i$ and then $\angle(x'_i,x'_j)\ge \max(\theta_i,\theta_j)\ge \theta$. In conclusion
$$\displaystyle |c_{ij}|\le \frac{1}{\sin(\theta)^2}$$$$\displaystyle |c_{ij}|\le \frac{\cos(\theta)}{\sin(\theta)^2}\quad\text{if } i\ne j$$ $$\displaystyle |c_{ii}|\le \frac{1}{\sin(\theta)^2}$$
and therefore if $\beta=\max\{\beta_i\}$ $$1=\sum_{i,j}c_{ij}\sin(\beta_i)\sin(\beta_j)\le \sin(\beta)^2\sum_{i,j}|c_{ij}|\le \frac{n^2\sin(\beta)^2}{\sin(\theta)^2}$$$$1=\sum_{i,j}c_{ij}\sin(\beta_i)\sin(\beta_j)\le \sin(\beta)^2\sum_{i,j}|c_{ij}|\le \sin(\beta)^2\frac{n+(n^2-n)\cos(\theta)}{\sin(\theta)^2}$$ $$\sin(\beta)\ge\frac{sin(\theta)}{\sqrt{n(1+(n-1)\cos(\theta))}}$$
whichThis also implies $\sin(\beta)\ge \sin(\theta)/n$, andthe good $\beta\ge \theta/n$ by monotonicity of(as $\sin(x)/x$ in$\theta\rightarrow 0$) first order approximation $[0,\pi/2]$$\sin(\beta)\ge \sin(\theta)/n$. $\quad\blacksquare$
I verified empirically that $\frac{sin(\theta)}{\sqrt{n(1+(n-1)\cos(\theta))}}\ge \sin(\frac{\theta}{n})$, which would imply $\beta\ge\frac{\theta}{n}$ too, but I'll leave the proof of that as an exercise, if true.
The example in my other answer shows this result isled me to conjecture that the best possible, in first order result should be $$\sin(\beta)\ge \sin(\theta)\sqrt{\frac{(n-1)\tan(\theta)^2+n}{n((n-1)\tan(\theta)^2+n^2)}}$$
which is not much stonger than the claim, as $\theta\rightarrow 0$but probably much harder to prove.