Well, n is certainly bound above by N-1, and that bound can be attained. It is a nice exercise to show that, passing to an algebraic closure as necessary, that for an NxN matrix Tr(X^r)=0 for all r from 1 to N inclusive implies that X is nilpotent - this is becuase these polys are a basis for the symmetric polys in the N eigenvalues (counted with multiplicities) of X, and if they are all zero then so is the product of all the eigenvalues as that is another symmetric poly, which in turn implies one e-value is zero, and by induction all e-values are zero, and X is nilpotent. Since unitary matrices are not nilpotent that puts N as the strict upper bound on n in your question. Certainly 1 is attainable for 2x2 matrices, and it is easy to see that you can get n=N-1 for N prime. I haven't checked your example, but I see no reason not believe you haven't checked it.