The determinant and the trace are two quite different beasts, little relation can be found among them.
If the matrix is not only symmetric (hermitic) but also _positive semi-definite_ , then its eigenvalues are real and non-negative. Hence, given the properties#Eigenvalue_relationships) ${\rm tr}(M)=\sum \lambda_i$ and ${\rm det}(M)=\prod \lambda_i$, and recalling the AM GM inequality, we get the following (probably not very useful) inequality:
$$\frac{{\rm tr}(M)}{n} \ge {\rm det}(M)^{1/n}$$
(equality holds iff $M = \lambda I$ for some $\lambda \ge 0$)
Much more interesting/insightful/useful are the answers by Owen Sizemore and Rodrigo de Azevedo.