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.
On
(Of course the trivial group acts freely on all spheres! Let's assume $G$ is finite and nontrivial.)
Since $G$ is finite, the action is certainly properly discontinuous, and since the action is given to be free, it follows that the quotient $q: X \rightarrow X/G$ is a finite covering map. Recalling that in any $n$-sheeted covering map $q: X \rightarrow Y$ we have $\chi(X) = n \chi(Y)$ and that the Euler characteristic must be an integer, we are almost done.
Do you know the definition of degree? You can arguing as follows. Suppose a group G act on 2n-sphere. Then there is a homomorphism from G to {1,-1} which sends an element g to the degree of g-action. But if g is not identity, then g(x) is not x, thus the g-action is homotopic to the antipodal map. And in sphere of even dimension, the antipodal map is of degree -1(because it reverse the orientation). Let's see what do we have:(1) a homomorphism p:G->$C_2$; (2)p(g)=-1 if g is not identity. So now you can conclude that G actually has only two element.(So I don't understand why you assume that G is a finite group.)