When $n$ is even, it is easy to classify groups which act freely on $S^n$ using degree theory: if $G$ acts on $S^n$, then associating to each element $g \in G$ the degree of the map obtained from multiplication by $g$, one gets a map $d : G \to \{\pm 1\}$. It is easy to verify this is a homomorphism. If $G$ acts freely, multiplication by any element $g$ is a fixed point free map, thus $d(g) = (-1)^{n+1} = -1$, making $d$ injective. The only nontrivial group which injects into $\Bbb Z/2$ is itself, so we're done.
However, a lot of groups act freely on $S^n$ for odd $n$. For example, $\Bbb Z/p$ acts on $S^3$ freely for all prime $p$ (so called lens space action). What do we know about such groups? Is it possible to classify them?
If $G$ is a finite group acting freely on $S^3$, then as free action of finite groups on Hausdorff spaces is properly discontinuous, quotient map $S^3 \to S^3/G$ is a covering projection. Thus $S^3/G$ is a closed 3-manifold with fundamental group $G$, hence $G$ is a closed 3-manifold group. On the other hand, if $G$ is a closed 3-manifold group, let $M$ be that manifold and $G$ must act on $\tilde{M}$ freely. But $\tilde{M}$ is a simply connected closed 3-manifold hence homeomorphic to $S^3$ by Poincare conjecture, so $G$ must act on $S^3$ freely.
Thus, finite groups acting freely on $S^3$ are precisely the finite closed 3-manifold groups. But what about infinite groups? Given an infinite group, how can we tell if it acts on $S^3$ or not? More generally, what about groups acting freely on $S^n$ for some fixed odd $n > 1$?
[edit] I am only interested in actions of discrete groups. Also, any sort of general remark (long enough to not fit as a comment) or partial answers (like the answers below) are welcome to me, you can post them as answers.