Given a finite group $G$, e.g. $G=S_3$, is there a finite dimensional manifold $M$ such that $\pi_1(M)=G$?

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It is known that the standard Eilenberg-Maclane space $K(G,1)$ is uniquely determined by any given group $G$. However, if $G$ contains an element of finite order, then every $K(G,1)$ CW complex must be infinite dimensional.

We are interested in the situation of $G$ is finite and non-abelian, and we do not ask additional condition on its universal covering space (in other words, $\pi_i(M)=0$, for $i>1$).

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The following article by Rubinstein constructs such manifolds, asked in the title:

On $3$-manifolds that have finite fundamental group and contain Klein bottles.

For $4$-manifolds, see this MO-question.