If $G$ is a discrete group, and if $BG$ can be represented by a (finite dimensional) manifold, is it always true that $EG$ can be chosen to be $\mathbb{R}^n$, for some $n?$
I'm guessing the answer is no, but the fact that the Whitehead manifold satisfies $W\times\mathbb{R}\cong \mathbb{R}^4$ does make me wonder.
A $k$-manifold $M$ represents $BG$ if and only if $\pi_1(M) \cong G$ and $\widetilde{M}$, the universal cover of $M$, is contractible ($\widetilde{M}$ represents $EG$). If $M$ has boundary, we can choose a different representative which doesn't have boundary (namely $M\setminus \partial M$ which is homotopy equivalent to $M$). From now on, we will suppose $M$ has no boundary.
Note that $\widetilde{M}$ is homotopy equivalent to $\mathbb{R}^k$. If $k \leq 2$, then in fact $\widetilde{M}$ is homeomorphic to $\mathbb{R}^k$ by the classification of such manifolds. This is no longer true when $k > 2$, e.g. if $G$ is trivial, then $BG = EG$ is represented by the Whitehead manifold $W$.
We say a topological space $X$ is simply connected at infinity if for any compact subset $C \subset X$, there is a compact set $D$ with $C \subset D \subset X$ such that the induced map $\pi_1(X\setminus D) \to \pi_1(X\setminus C)$ is the zero map. This property is invariant under homeomorphisms.
Example: $\mathbb{R}^k$ for $k > 2$ is simply connected at infinity. To see this, just take $D$ to be a large closed ball containing $C$, then $\pi_1(\mathbb{R}^k\setminus D) \cong \pi_1(S^{k-1}) = 0$, so the map $\pi_1(\mathbb{R}^k\setminus D) \to \pi_1(\mathbb{R}^k\setminus C)$ is the zero map.
Non-Example: The Whitehead manifold $W$.
This is how one sees that $W$ is not homeomorphic to $\mathbb{R}^3$. When $k > 2$, all such examples take this form, i.e. an open contractible $k$-manifold with $k > 2$ is homeomorphic to $\mathbb{R}^k$ if and only if it is simply connected at infinity, see this answer.
Now consider $N := M\times\mathbb{R}$. We have $\pi_1(N) \cong \pi_1(M) \cong G$ and $\widetilde{N} = \widetilde{M}\times\mathbb{R}$ so $\widetilde{N}$ is contractible and hence $N$ also represents $BG$. Furthermore, $\widetilde{N}$ is simply connected at infinity, see Proposition $2.2$ of The Piecewise Linear Structure of Euclidean Space by Stallings. Therefore $\widetilde{N}$ is homeomorphic to $\mathbb{R}^{k+1}$ - note, the same trick shows why $W\times\mathbb{R}$ is homeomorphic to $\mathbb{R}^4$.
That is, if $BG$ can be represented by a finite-dimensional manifold $M$, then it can be represented by a finite-dimensional manifold $N$ such that $\widetilde{N}$, the corresponding representative of $EG$, is homeomorphic to $\mathbb{R}^n$ (namely $N = M\times\mathbb{R}$ and $n = \dim M + 1$).