The domain of the Haar measure of the $n$-dimensional Torus

The answer is yes. For the natural projection $\pi:\mathbb{R}^n\to\mathbb{T}^n$, the measurable subset $\Sigma=[0,1[^n\subseteq \mathbb{R}^n$ is a strict fundamental domain (i.e. $\pi|_\Sigma:\Sigma\to \mathbb{T}^n$ is a bimeasurable bijection).

Let $B\subseteq \mathbb{T}^n$ be a subset such that $\pi^{-1}(B)\in\mathfrak{B}({\mathbb{R}^n})$. Then

$$B=\pi|_\Sigma((\pi|_\Sigma)^{-1}(B))= \pi|_\Sigma(\Sigma\cap \pi^{-1}(B)).$$

Since $\Sigma$ and $\pi$ are measurable, so is $\Sigma\cap \pi^{-1}(B)$. Since $(\pi|_\Sigma)^{-1}$ is measurable so is $B$. Thus $\pi_\ast(\mathfrak{B}(\mathbb{R}^n))\leq \mathfrak{B}(\mathbb{T}^n)$.

See https://mathoverflow.net/q/50541/66883 for a more general version of this argument. Also relevant is Characterization of the Borel $\sigma$-algebra on a topological quotient space.