Denote the given map by $\phi: \mathbb C/\Lambda \to \mathbb P^{n-1}$.
Let $H$ be a hyperplane in $\mathbb P^{n-1}$, cut out by the equation $a_1 X_1 + \cdots + a_n X_n = 0$. The preimage $\phi^{-1}(H)$ is then equal to the subset of points of $\mathbb C/\Lambda$ satisfying the equation $a_1 f_1 + \cdots a_n f_n = 0$.
First of all, since $n \geq 3$, Riemann--Roch says that given any two points $u$ and $v$ of $\mathbb C/\Lambda$, we may choose the $a_i$ so that $u$ lies in $\phi^{-1}(H)$ while $v$ does not. This implies that $\phi$ is injective. (Note that here and below I am being a little careless about analyzing what happens at the point $u = 0$ here, where we have to modify the formula for $\phi$; I'll leave these details to you.)
In fact, Riemmann--Roch also implies that we can choose $H$ so that $a_1 f_1 + \cdots a_n f_n$ vanishes precisely to first order at any given point $u$. This show that the image of $\phi$ is smooth at $\phi(u)$; since $u$ was arbitrary, we see that the image of $\phi$ is smooth, and hence that $\phi$ is an embedding onto its image. (I have just sketched the verification that $\phi$ "separates points and tangent vectors", which is the standard condition to check that a map defined in the manner of $\phi$ is a projective embedding; this is discussed e.g. in Hartshorne Chapter 2, Section 7, and also --- in the specific case of curves --- in the part of Chapter 4 that discusses projective embeddings of curves.)
Since $\mathbb C/\Lambda$ is compact, its image is a closed analytic submanifold of $\mathbb P^{n-1}$, which is then necessarily algebraic (by Chow's theorem if you like, although you could also prove this directly, by showing ----- by Riemann--Roch or by direct elementary arguments --- that the $f_i$ necessarily satisfy lots of algebraic relations among themselves, indeed so many that the image of $\phi$ is cut out by algebraic equations. A more systematic way to describe the situation is to note that $\mathbb C/\Lambda$ has a unique structure of a projective algebraic curve --- e.g. the one coming from the Weierstrass equation arising from the case $n = 3$ --- and with respect to this structure, the $f_i$ are rational funtions; the map $\phi$ is then a morphism of algebraic varities from $\mathbb C/\Lambda$ to $\mathbb P^{n-1}$; since its source is projective, its image is necessarily a closed algebraic subvariety.)
Now all that remains is to compute the degree of the image of $\phi$. For this, we have to compute the number of intersection points of this image with a generic hyperplane $H$. That is, we need to determine the size of $\phi^{-1}(H)$. But this is the number of zeroes of $a_1 f_1 + \cdots + a_n f_n$. Since this function has a pole of order $n$ (at least if $a_n \neq 0$, which is the generic situation), it has $n$ zeroes, which will be distinct genericaly, and so $\phi^{-1}(H)$ is generically of order $n$.
Thus the image of $\phi$ has degree $n$.
If you're not used to these sort of computations, you may want to look at some of the examples and discussion of Chapter 4 of Hartshorne. What I just did above is the standard way of analyzing projective embeddings of curves, and in that sense it is easy (once you know how it goes).