According with Mumford the answer is yes, but there are some obscure points in the proof.
We know that there exists a very ample line bundle on $A$ since every abelian variety is projective. Hence, via some corollaries of the Theorem of the Cube, we get an ample line bundle whose restriction to the kernel of $n_A$ is both the trivial bundle and very ample. This should imply that the dimension of $\ker n_A$ is $0$.
Furthermore, Mumford says that the fact that the dimension of the kernel is $0$ implies the surjectivity of $n_A$.
Is anybody able to clarify these two points to me?
Thanks in advance!