Let $\alpha_1, \dots, \alpha_n$ be unit vectors in some vector space $V = R^d$. For any permutation $\pi: [n] \rightarrow [n]$, we can form the Gram-Schmidt orthogonal bases $\beta_{\pi,1}, \dots, \beta_{\pi,n}$ as $\beta_{\pi, 1} = \alpha_{\pi(1)}, \beta_{\pi,2} = \alpha_{\pi(2)} - (\alpha_{\pi(2)} . \beta_{\pi,1}) \beta_{\pi,1}$ etc.
Now, for any permutation $\pi$ define the set $X_{\pi} = \{ x \in V \mid (x.\beta_{\pi,i})^2 \geq \lambda \text{ for $i = 1, \dots, n$} \}$.
And define the set $X = \cup_{\pi} X_{\pi}$.
What does this set look like? Can one bound its volume?
Obviously, you can sum the volume of $X_{\pi}$ over $\pi \in S_n$. However, intuitively, it seems that the worst case for the volume of $X_{\pi}$ would be when the $\alpha$ vectors are orthonormal --- otherwise, the $\beta$ vectors become smaller. But in that case, $X_{\pi}$ does not depend on $\pi$ at all. So it seems like you should be able to avoid this extra factor of $n!$
