Is the projection from $\mathbb R^n$ locally isometric to the flat torus? I’m reading a problem asking to introduce a Riemannian metric on $T^n$ in such a way that the projection $\pi(x_1,...,x_n)=(\exp(ix_1),...,\exp(ix_n))$ from $\mathbb R^n$ to $T^n$ is a local isometry. Does the flat metric apply here?

By definition, flat torus is defined as the product of n copies of the Riemannian manifold $S^1$ with its metric induced by $\mathbb R$.
Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.
Notice that the natural projection written above maps a point of $R^n$ to a point in $T^n$. This may seem odd at first but notice how the coordinate curves of $R^n$ get mapped to $T^n$. Because $e^{ix}$ is a point in a complex circle, the projection essentially wraps each axis of $R^n$ into a circle of the torus. This maps coordinate curves into coordinate curves. Since the projection is evidently the cartesian product of a number of complex circles (which are the same as $S^1$) you can introduce the product metric the same as if you were working with real numbers exclusively. Since you can take the vectors to be coordinate curves locally, the product metric is a local isometry.