Not every Hermitian operator is an observable. The identity on a space is certainly Hermitian, but it is no observable in any physical sense - its eigenvalue is $1$, and that's it. The eigenvalue of scalar operators gives you precisely zero information about the state considered, and usually, we would like an observable to at least give us some information about the state we measure.
This is a subtlety that is usually glossed over in most courses, but if you think about canonical quantization, it is quite clear that not every Hermitian operator on the space of states will be induced by an observable on the classical phase space.
The density matrix is similar - yuggib's interpretation is correct, but this is quite unlike our usual observables, where the eigenvalues of the observable would correspond to some (classical) property of the state being measured. I'm reluctant to unequivocally pronounce it an "observable" or not on these grounds.