For $n=1, 2$ the answer is "yes" since the group is virtually free, for $n\ge 3$ the answer is not known (an open problem).
Edit. In fact even for two $n\times n$-matrices the problem is open. Moreover the solution of the following ``easier" problem is not known: for which algebraic integers $\lambda$ the matrices $\left(\begin{array}{ll} 1 & 2\\ 0 & 1 \end{array}\right)$ and $\left(\begin{array}{ll} 1 & 0\\ \lambda & 1 \end{array}\right)$ generate a free group (see this paper, for example). The fact that this problem is easier follows from the trivial observation that the group generated by these two matrices is isomorphic to some effectively computable group of $n\times n$-integer matrices for some $n\ge 2$ (depending on the degree of the algebraic number $\lambda$).
