A simple estimate which is often useful is that, if $A$ and $B$ are Hermitian matrices with eigenvalues $a_1 > a_2 > \ldots > a_n$ and $b_1 > b_2 > \ldots > b_n$ and the eigenvalues of the sum are $c_1 > c_2 > \ldots > c_n$, then
$$
c_{i+j-1} \le a_i + b_j \quad\text{and}\quad c_{n-i-j} \ge a_{n-i} + b_{n-j}.
$$
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 $(a_1, \ldots, a_n, b_1, \ldots, b_n, c_1, \ldots, c_n)$ which occur as eigenvalues of $(A,B,C)$ with $A+B=C$ are dense in the hyperplane $\sum a_i + \sum b_i = \sum c_i$.