The standard metric on $\mathbb{R}^n$, namely $g = \sum_{i=1}^ndx^i\otimes dx^i$, is invariant under translations, i.e. for $f: \mathbb{R}^n \to \mathbb{R}^n$ of the form $f(x) = x + a$ where $a = (a^1, \dots, a^n) \in \mathbb{R}^n$, we have
\begin{align*}
f^*g &= f^*\left(\sum_{i=1}^ndx^i\otimes dx^i\right)\\
&= \sum_{i=1}^n f^*dx^i\otimes f^*dx^i\\
&= \sum_{i=1}^nd(x^i\circ f)\otimes d(x^i\circ f)\\
&= \sum_{i=1}^nd(x^i + a^i)\otimes d(x^i+a^i)\\
&= \sum_{i=1}^ndx^i\otimes dx^i\\
&= g.
\end{align*}
In particular, $g$ is invariant under translations by elements of the lattice $\mathbb{Z}^n$.
As $\pi : \mathbb{R}^n \to T^n$ is a smooth covering map, it is a local diffeomorphism. So given $q \in T^n$ and $p \in \pi^{-1}(q)$, $(\pi_*)_p : T_p\mathbb{R}^n \to T_qT^n$ is an isomorphism. Now for $v, w \in T_qT^n$, define
$$\hat{g}_q(v, w) := g_p((\pi_*)_p^{-1}(v), (\pi_*)_p^{-1}(w)).$$
To see this definition is well-defined (i.e. independent of the choice of $p$), note that if $p' \in \pi^{-1}(q)$, there is $a \in \mathbb{Z}^n$ such that $p' = p + a$. Letting $f : \mathbb{R}^n \to \mathbb{R}^n$ denote translation by $a$, we have $f(p) = p'$ and $\pi\circ f = \pi$, so $(\pi_*)_p = ((\pi\circ f)_*)_p = (\pi_*)_{f(p)}\circ(f_*)_p = (\pi_*)_{p'}\circ(f_*)_p$. Therefore,
\begin{align*}
g_{p'}((\pi_*)_{p'}^{-1}(v), (\pi_*)_{p'}^{-1}(w)) &= g_{f(p)}((f_*)_p\circ(f_*)_p^{-1}\circ(\pi_*)_{p'}^{-1}(v), (f_*)_p\circ(f_*)_p^{-1}\circ(\pi_*)_{p'}^{-1}(w))\\
&= g_{f(p)}((f_*)_p\circ((\pi_*)_{p'}\circ(f_*)_p)^{-1}(v), (f_*)_p\circ((\pi_*)_{p'}\circ(f_*)_p)^{-1}(w))\\
&= g_{f(p)}((f_*)_p\circ(\pi_*)_p^{-1}(v), (f_*)_p\circ(\pi_*)_p^{-1}(w))\\
&= (f^*g)_p((\pi_*)_p^{-1}(v), (\pi_*)_p^{-1}(w))\\
&= g_p((\pi_*)_p^{-1}(v), (\pi_*)_p^{-1}(w))\\
&= \hat{g}_q(v, w).
\end{align*}
So $\hat{g}$ is a well-defined Riemannian metric on $T^n$. By construction, $\pi^*\hat{g} = g$; that is, $\pi$ is a local isometry.
The exact same argument can be used to show that if $\pi : M \to N$ is a smooth covering map, and $g$ is a Riemannian metric on $M$ which is invariant under the deck transformations of $\pi$, then it descends to a Riemannian metric $\hat{g}$ on $N$, and $\pi^*\hat{g} = g$; that is, $\pi$ is a local isometry.
An attempt of proof, please review carefully and tell me your opinion. Thanks and kind regards.
Recall that the $n$-torus $T^n$ is defined as the quotient $\Bbb{R}^n/G$ where $G$ is the group of integer translations in $\Bbb{R}^n$. It can be identified with the product of $n$ circles:
$$
T^n = \underbrace{S^1 \times \ldots \times S^n}_{n \text{ times}}.
$$
We can then define a natural projection $\pi: \Bbb{R}^n \longrightarrow T^n$ given by
$$
\pi(x) = (e^{ix_1}, \ldots, e^{ix_n}), \quad x = (x_1, \ldots, x_n) \in \Bbb{R}^n.
$$
For $u = (u_1, \ldots, u_n) \in T_x\Bbb{R}^n = \Bbb{R}^n$ we have
$$
d\pi_x(u) = J(x)u = i(u_1 e^{i x_1}, \ldots, u_n e^{i x_n}).
$$
where $J(x)$ is the Jacobian matrix of $\pi$ at $x$ given by
$$
J(x) = \begin{bmatrix}
ie^{ix_1} & 0 & \cdots & 0 \\
0 & ie^{i x_2} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & i e^{i x_n}
\end{bmatrix}.
$$
We can just define
$$
\langle d\pi_x(u), d \pi_x(v) \rangle_{\pi(x)} = \langle u, v \rangle, \quad u, v \in T_x\Bbb{R}^n = \Bbb{R}^n
$$
where $\langle \cdot, \cdot \rangle$ denotes the inner product of the Euclidean space. To get a local isometry, it suffices to restrict $\pi$ to a neighborhood of $x$ such that it is a diffeomorphism.
We now show that the identity map $i: \Bbb{R}^n/G \longrightarrow T^n$ is an isometry, that is, we show that the two metrics are the same. Let $u,v \in T_pT^n$. Then
\begin{align*}
\langle u, v \rangle_{(e^{i x_1}, \ldots, e^{i x_n})} & = \sum_1^n \langle d \pi_j (u), d \pi_j(v) \rangle_{e^{i x_j}} \\
& = \sum_1^n \langle u_j e^{i (x_j + \pi/2)}, v_j e^{i (x_j + \pi/2)} \rangle_{e^{i x_j}} \\
& = \sum_1^n u_j v_j \\
& = \langle u, v \rangle,
\end{align*}
which completes the proof.
Best Answer
Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.