Here is a sketch of a purely algebraic proof based on the theory developed in Chapter 3 of Mumford's "Abelian Varieties".

Let $L$ be a nef line bundle on the abelian variety $A$ of dimension $g$. If $K(L)$ is finite, then the index of $L$ is a well defined integer between $0$ and $g$ and we need to show that $g=0$. By the theorem on p. 155, the index is the number of positive roots of the polynomial $P(n) = \chi(L \otimes M^n)$, where $M$ is any ample line bundle. Using Riemann-Roch and the nefness of $L$ one sees that all the coefficients of the polynomial are non-negative so it has no positive roots.

We now reduce the general case to the one above. We may assume, by replacing $L$ by $L \otimes (-1)^*L$, that $L$ is symmetric. If $K(L)$ is not finite, i.e. $T_x^*L \cong L$ for $x$ in a positive dimensional subvariety $B$ of $A$ we claim that $L^{2}$ descends to a line bundle $L'$ on $A/B$. This follows as we can get descent data for $L^2$ by choosing symmetric isomorphisms $T_x^*L \cong L$ for all $x \in B$. (Note that one cannot always descend $L$ as shown by the example of a line bundle of order $2$ in $Pic(A)$.)

Since $L$ is nef it follows that $L'$ is also nef, so it is numerically equivalent to an effective line bundle by induction.

(In the sketch above, the class of $L$ modulo numerical equivalence is replaced with a multiple. One can perhaps avoid this by an additional argument.)