Here is a theorem due to Armstrong:

A group $G$ of homeomorphisms of a topological space $X $is called discontinuous if (1) the stabilizer of each point of $X$ is finite, and (2) each point $\chi \in X$ has a neighborhood $U$ such that any element of $G$ not in the stabilizer of $\chi$ maps $U$ outside itself. Theorem: Let $G$ be a discontinuous group of homeomorphisms of a path connected, simply connected, locally compact metric space $X,$ and let $H$ be the normal subgroup of $G$ generated by those elements which have fixed points. Then the fundamental group of the orbit space $X/G$ is isomorphic to the factor group $G/H.$

(the above is actually the math review of

Armstrong, M. A.
The fundamental group of the orbit space of a discontinuous group.
Proc. Cambridge Philos. Soc. 64 1968 299–30
)

So, in order for the (underlying topological space of the) quotient of $\mathbb{R}^n$ to be simply connected, it must be generated by elements with fixed points (and vice versa). Of course, a quotient of $\mathbb{T}^n$ is also a quotient of $\mathbb{R}^n,$ so this applies to your question. On the other hand, simply connected does not a sphere make, but figuring out the higher homology groups of the quotient should be relatively easy, and if they are all trivial, then the quotient is a sphere by the Poincare conjecture.

(By the way, the reason I know of Armstrong's theorem is that it was used by Colin Maclachlan to show that the (underlying topological space of the) moduli space of Riemann surfaces is simply connected (since the mapping class group acts discontinuously, and is generated by torsion).

**A little example** Of course, in two dimension, the standard example is the quotient of the torus $\mathbb{T}^2$ by the elliptic involution, which is the sphere with four distinguishe (Weierstrass) points.

**A bigger example** If I understand correctly, any projective variety has a branch covering map to $\mathbb{P}^n.$ (see, e.g., https://math.stackexchange.com/questions/679768/projective-noether-normalization). Since there are plenty of projective tori running around, you always have branched cover maps to $\mathbb{CP}^n.$

**Three dimensions**
This case was completely analyzed by Bill Dunbar in his thesis. It seems that most torus quotients with singular locus have $S^3$ as their underlying topological space.

**more examples**
Will Sawin's example for $n=2$ gives $S^1.$ For $n=3$ it gives $S^2,$ with three singular points of cone angle $2\pi/3,$ so a doubled regular triangle. In dimension $3,$ the nicest example of a branched cover of $S^3$ by $T^3$ is the orbifold whose branch locus is the Borromean rings (see page 81 in this 2008 paper by Montesinos, which is highly recommended in general), with branching index $2.$ This might be the Sawin example for $n=4,$ but I am not completely certain.

**Is it a manifold?** In a comment discussion with Will Sawin, we agreed that a key part of the question is whether the underlying topological space is actually a manifold. This is equivalent to the quotient of the link of every branch point being a topological sphere (it is *always* a rational homology sphere). Now, the dimension of the link is $n-1,$ so since every rational homology sphere in dimensions $0, 1, 2$ is an actual sphere, the question is vacuous for $n=1, 2, 3.$ For $n=4,$ it becomes nontrivial. Luckily, spherical $3$-orbifolds have been studied by Bill Dunbar. Most are, indeed, topological spheres, see:

MR1118824 (94c:57022) Reviewed
Dunbar, William D.(1-PASE)
Nonfibering spherical 3-orbifolds.
Trans. Amer. Math. Soc. 341 (1994), no. 1, 121–142.

**but there is more**
Notice that every element of $SO(2n-1)$ has fixed points. Therefore, for a five dimensional orbifold (assuming the point stabilizer action is linear, which is certainly true for a crystallographic action on $\mathbb{T}^5,$ the vertex links are simply connected. They are not necessarily manifolds, but if they are, they are simply connected rational homology spheres of dimension 4. Does this mean they are spheres? This merits anther question...