Every holomorphic map from Complex projective line to complex torus is constant + generalization. Here is the result:

Every holomorphic map from $\mathbb{C}P^1$ into a complex torus is constant.

And what about a holomorphic map from $\mathbb{C}P^n$ into a complex torus?
I have no idea at the moment.. This is the exercise 2.1.4. from Huybrechts book, "Complex geometry" 
 A: Any holomorphic mapping from the complex $n$-dimensional complex projective space into a complex $m$-torus is constant.
To see this, recall the following result from the theory of covering spaces. (It essentially is Theorem 4.17. in Forster's Lectures on Riemann Surfaces.)
Theorem. Let $X$ and $Y$ be Hausdorff topological spaces. Let $Z$ be a simply connected, locally path connected and path connected topological space. Let $\pi\colon X\rightarrow Y$ be a (not necessarily surjective) covering map. Then any continuous map $f\colon Z\rightarrow Y$ lifts to a continuous map $\hat{f}\colon Z\rightarrow X$ along $\pi$, i.e. we have $f=\pi\circ\hat{f}$.
Now, let $\Gamma\subsetneq \mathbb{C}^m$ be a lattice in $\mathbb{C}^m$. Let $f\colon \mathbb{CP}^n\rightarrow \mathbb{C}^m/\Gamma$ be a holomorphic mapping into the complex $m$-torus $\mathbb{C}^m/\Gamma$.
To apply above theorem to your problem, first check that the quotient map $\pi\colon \mathbb{C}^m\rightarrow \mathbb{C}^m/\Gamma$ is a covering map. For instance, you could show that the group action $\Gamma\times \mathbb{C}^m\rightarrow \mathbb{C}^m;(k,z)\mapsto k+z$ is properly discontinuous. Next, note that the complex $n$-dimensional complex projective space $\mathbb{CP}^n$ is locally path connected (it is even a complex $n$-dimensional manifold) and path connected (it is the image of a path connected space under a continuous map). Additionally, convince yourself that $\mathbb{CP}^n$ is simply connected.
Now, since holomorphic mappings are in particular continuous, by above theorem, there exists a continuous lift $\hat{f}\colon \mathbb{CP}^n\rightarrow \mathbb{C}^m$ of $f$ along $\pi$. We convince ourselves that $\hat{f}$ is a holomorphic mapping. To see this, let $z\in\mathbb{CP}^n$. Since $\pi$ is a covering map, it is a homeomorphism near $\hat{f}(z)$. Thus, near $\hat{f}(z)$ the quotient map $\pi$ has a local inverse, say $\pi^{-1}$. Since $\pi$ is holomorphic, this inverse is holomorphic. Thus, near $z$ the map $\hat{f}$ is of the form $\pi^{-1}\circ f$. Said differently, locally the map $\hat{f}$ is the composition of holomorphic mappings. Hence, the map $\hat{f}$ is holomorphic.
Next, assume for a contradiction that $\hat{f}$ was non-constant. Then there would exist $i\in\{1,\ldots m\}$ such that the holomorphic function $p_i\circ\hat{f}\colon \mathbb{CP}^n\rightarrow \mathbb{C}$ was non-constant. Here, the function $p_i\colon \mathbb{C}^m\rightarrow \mathbb{C}$ denotes the canonical projection onto the $i$-th coordinate. By the Open mapping theorem for holomorphic functions of several variables (Theorem 3.1. in Laurent-Thiébaut's Holomorphic Function Theory in Several Variables), the function $p_i\circ\hat{f}$ would be an open mapping. Thus, since $\mathbb{CP}^n$ is compact and $p_i\circ\hat{f}$ is continuous, the image of $p_i\circ\hat{f}$ would be closed and open in $\mathbb{C}$. Since $\mathbb{C}$ is connected, this would imply that $p_i\circ\hat{f}$ is surjective. This contradicts the fact that $\mathbb{C}$ is not compact. We conclude that $\hat{f}$, and consequently $f$, is constant.
