As Tom pointed out, the case that $n=0$ is trivial.

For $n=1$, if the group acts properly discontinously and cocompactly, then the group will have two ends, and therefore is isomorphic to $\mathbb{Z}$ by a result of Wall (Lemma 4.1)

One may see that any such properly discontinuous non-trivial torsion-free action must be cocompact. Suppose there is $g\in G$ such that $g(M\times\{0\}) \cap (M\times \{0\})=\emptyset$. Then by homology there is a compact submanifold $K\subset M\times \mathbb{R}$ such that $\partial K= (M\times\{0\}) \cup g(M\times\{0\})$. Then one sees that $K$ forms a fundamental domain for the action of $\langle g\rangle$ on $M\times \mathbb{R}$ by properness, and thus the action of $G$ is cocompact.

Otherwise, one has $g(M\times\{0\})\cap (M\times\{0\})\neq \emptyset$ for all $g\in G$. This violates proper discontinuity, since the preimage of the compact set $(M\times\{0\})\times (M\times\{0\})$ under the map $G\times (M\times\mathbb{R})\to (M\times\mathbb{R})\times (M\times\mathbb{R})$ is not compact.

As Tom pointed out, the case that $n=0$ is trivial.

For $n=1$, if the group acts properly discontinously and cocompactly, then the group will have two ends, and therefore is isomorphic to $\mathbb{Z}$ by a result of Wall (Lemma 4.1)

One may see that any such properly discontinuous non-trivial torsion-free action must be cocompact. Suppose there is $g\in G$ such that $g(M\times\{0\}) \cap (M\times \{0\})=\emptyset$. Then by homology there is a compact submanifold $K\subset M\times \mathbb{R}$ such that $\partial K= (M\times\{0\}) \cup g(M\times\{0\})$. Then one sees that $K$ forms a fundamental domain for the action of $\langle g\rangle$ on $M\times \mathbb{R}$ by properness, and thus the action of $G$ is cocompact.

Otherwise, one has $g(M\times\{0\})\cap (M\times\{0\})\neq \emptyset$ for all $g\in G$. This violates proper discontinuity, since the preimage of the compact set $(M\times\{0\})\times (M\times\{0\})$ under the map $G\times (M\times\mathbb{R})\to (M\times\mathbb{R})\times (M\times\mathbb{R})$ is not compact.

Another case one can deal with partially is for $M=S^1, n=2$. Then if $G$ acts cocompactly, then $G$ is as surface group by Theorem 1.2 of Maillot (although this theorem is attributed to Geoff Mess in an earlier unpublished preprint). This also follows now from the geometrization theorem now, but note that Maillot also partially treats the non-cocompact action case with some extra hypotheses.

As Tom pointed out, the case that $n=0$ is trivial.

For $n=1$, if the group acts properly discontinously and cocompactly, then the group will have two ends, and therefore is isomorphic to $\mathbb{Z}$ by a result of Wall (Lemma 4.1). I'm not sure what happens in the case the action is not cocompact, although I don't know an example off the top.

As Tom pointed out, the case that $n=0$ is trivial.

For $n=1$, if the group acts properly discontinously and cocompactly, then the group will have two ends, and therefore is isomorphic to $\mathbb{Z}$ by a result of Wall (Lemma 4.1)

One may see that any such properly discontinuous non-trivial torsion-free action must be cocompact. Suppose there is $g\in G$ such that $g(M\times\{0\}) \cap (M\times \{0\})=\emptyset$. Then by homology there is a compact submanifold $K\subset M\times \mathbb{R}$ such that $\partial K= (M\times\{0\}) \cup g(M\times\{0\})$. Then one sees that $K$ forms a fundamental domain for the action of $\langle g\rangle$ on $M\times \mathbb{R}$ by properness, and thus the action of $G$ is cocompact.

Otherwise, one has $g(M\times\{0\})\cap (M\times\{0\})\neq \emptyset$ for all $g\in G$. This violates proper discontinuity, since the preimage of the compact set $(M\times\{0\})\times (M\times\{0\})$ under the map $G\times (M\times\mathbb{R})\to (M\times\mathbb{R})\times (M\times\mathbb{R})$ is not compact.