There are many closed manifolds with universal cover homotopy equivalent to $\mathbb{R}^n$, they are precisely the closed aspherical manifolds. There are also many closed smooth manifolds with universal cover diffeomorphic to $\mathbb{R}^n$, e.g. those which admit a metric of non-positive curvature. If one weakens diffeomorphic to homeomorphic, then the only additional examples one could possibly obtain would be four-dimensional, but no such examples are known to exist, see this question.
If one considers $\mathbb{R}^n\setminus\{x\}$ with $n > 2$ instead as a universal cover, there are plenty of examples. Any quotient of $S^{n-1}$ will have universal cover $S^{n-1}$ which is homotopy equivalent to $\mathbb{R}^n\setminus\{x\}$. If one upgrades to diffeomorphism, then the product of a smooth quotient of $S^{n-1}$ with $S^1$ yields a suitable manifold. Unlike the case of $\mathbb{R}^n$, one can construct smooth manifolds with universal cover homeomorphic but not diffeomorphic to $\mathbb{R}^n\setminus\{x\}$. For example, for any exotic $(n-1)$-sphere $\Sigma$, the universal cover of $\Sigma\times S^1$ is diffeomorphic to $\Sigma\times\mathbb{R}$ which is homeomorphic to $S^{n-1}\times\mathbb{R}$, and hence $\mathbb{R}^n\setminus\{x\}$, but is not diffeomorphic to it, see these comments by Igor Belegradek.
What if we remove more than one point from $\mathbb{R}^n$?
Is there a closed manifold whose universal cover is homotopy equivalent/homeomorphic/diffeomorphic to $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$?
There are no such manifolds in dimension one or two, but I don't even know if such examples can arise in dimension three.
The space $\mathbb{R}^n\setminus\{x_1,\dots, x_k\}$ is homotopy equivalent to $\bigvee_{i=1}^kS^{n-1}$. If $M$ is a closed manifold with the given universal cover, one might hope that an analysis of the natural $\pi_1(M)$-action on $\pi_{n-1}(M) \cong \mathbb{Z}^k$ could provide some insight.
Best Answer
If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing point (together with the "infinity" of the one-point compactification of $\mathbb{R}^n$) is an end of the covering space, and these are all the ends. Therefore the universal cover has $k+1 \ge 3$ ends.
Freudenthal and Hopf proved that a finitely generated group has $0$, $1$, $2$, or infinitely many ends. The ends of the fundamental group biject with the ends of the universal cover, so the theorem contradicts our assumption.
I suspect (but am really not sure) that it may be possible to extend the argument to closed $n$-manifolds with universal cover only homotopy equivalent to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$. Perhaps it can be shown that the universal cover must have $k+1$ ends by thinking of the separation properties of representatives of $H_{n-1}(\bigvee_i S^{n-1})$.