I don't know how to show this.
Do I assume $G$ acts on $S^{2n}$ by homeomorphisms? Then, since $S^{2n}$ is Hausdorff I'd know $G$ acts freely and properly discontinuously, and since $\pi_1(S^{2n})={1}$ I'd have $\pi_1(X/G)\cong G$. But I'm not sure whether this is useful.