Homology group of $\tilde{P}$ as $\mathbb{Z}[\pi_1 (P)]$ when $P=\mathbb{RP}^n$

Question

Let $P=\mathbb{RP}^n$, $m\geq 2$. Let $\tilde{P}=S^n$ be the universal covering space of $P$. What is $H_n (\tilde{P})$ as $\mathbb{Z}[\pi_1 (P)]$?

I know that $\pi_1 (P)\cong \mathbb{Z}_2$ and $H_n (\tilde{P})\cong \mathbb{Z}$ as $\mathbb{Z}$-module. But how we can know it as $\mathbb{Z}[\pi_1 (P)]$? Is that a free module of rank 1?