This isn't exactly an answer to your question, but here's how I like to think about the fact that you quoted.

Let's assume for a minute that $n=2$, so that we can think of $X$ as the complement of the braid arrangement in $\mathbb{C}^k$. Let $G = \mathbb{Z}/2\mathbb{Z}$, which acts on $X$ by complex conjugation.

Replace $\mathbb{Q}$ with a field $F$ of characteristic $2$. The $G$-equivariant cohomology ring $H^*_G(X; F)$ is a free module over $H^*_G(pt; F) \cong F[x]$ with the property that specializing at $x=0$ gives $H^* (X; F)$ and specializing at $x=1$ gives $H^*(X^G;F)$.

Thus we have a family of $\Sigma_k$ representations over the $F$-affine line interpolating between $H^* (X; F)$ and $H^*(X^G; F)$. Since the category of $\Sigma_k$ representations is semisimple, these two representations have to be isomorphic. The fact that $H^*(X^G; F)$ is the regular representation is obvious.

This is a good way to see that $H^*(X; F)$ is isomorphic to the regular representation of $\Sigma_k$. I'm not sure how to modify this argument to get $H^* (X; \mathbb{Q})$. I'm also not sure if this will help you find a cyclic vector, since it does not give you an explicit isomorphism between $H^* (X; F)$ and $H^* (X^G; F)$.

By the way, for $n>2$ you can do something similar, where $G$ acts on $\mathbb{R}^n$ by negating the last $n-1$ coordinates.

Representation theory and homological stability(arxiv.org/abs/1008.1368), specifically section 4. – HJRW Nov 20 '10 at 18:11