A simple estimate which is often useful is that, if A and B are Hermitian matrices with eigenvalues a_1 > a_2 > ... > a_n and b_1 > b_2 > ... > b_n and the eigenvalues of the sum are c_1 > c_2 > ... > c_n, then

c_{i+j-1} ≤ a_i + b_j and c_{n-i-j} ≥ a_{n-i} + b_{n-j}.

The above conditions are necessary but not sufficient for A+B=C to have a solution; see the Knutson-Tao article if you want sufficient conditions.

If you do not impose that A and B are Hermitian then there are very few restrictions besides the trace being equal. More specifically, the 3n-tuples (a_1, ..., a_n, b_1, ..., b_n, c_1,...,c_n) which occur as eigenvalues of (A,B,C) with A+B=C are dense in the hyperplane \sum a_i + \sum b_i = \sum c_i.