Pauli principle for "Phonons" I'm reading in Feynman's "Statistical Mechanics" Chap. 6.4 about a system of $M$ interacting particles, they may be bosons or fermions. Let the hamiltonian be
$$ H=\sum_i^{3M}p_i^2+\sum_{ij}^{3M}U_{ij}q_iq_j,\tag{1} $$
with a symmetric matrix $U$ and apropriate definition of p and q to get rid of the constants. By diagonalizing, the Hamiltonian can be brought in the form
$$H=\frac{1}{2}\sum_i^{3M}(P_i^2+\omega_i^2Q_i^2),\tag{2}$$
with generalized coordinates $P_i$ and $Q_i$. Define the creation and anihilation operators $a_k=\frac{1}{\sqrt{2\hbar}}(\sqrt\omega_kQ_k+\frac{i} {\sqrt{\omega_k}}P_k)$ and $a_k^\dagger$.
The hamiltonian therefore takes the form
$$
H
=\sum_i^{3M}\hbar\omega_i\left(a_i^\dagger a_i+\frac{1}{2}\right)
=\sum_i^{3M}\hbar\omega_i\left(N_i+\frac{1}{2}\right).\tag{3} 
$$
The eigenstates are 
$$
|n_1n_2...n_{3M}\rangle
=\prod_i^{3M}\frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}}|\mathrm{vac}\rangle 
\tag{4}
$$
$$
H|n_1n_2...n_{3M}\rangle
=\sum_i^{3M}\hbar\omega\left(n_i+\frac{1}{2}\right) |n_1n_2...n_{3M}\rangle, 
\tag{5} 
$$
and they are interpreted as $n_1$ "phonons" in the first mode, $n_2$ in the second and so on. 
Now to the questions:


*

*How do we have to apply the Pauli principle here?  I suppose the states are tensor products of "single particle" states $|n_1n_2...n_{3M}\rangle:=|n_1\rangle|n_2\rangle...|n_{3M}\rangle$, but in that case they would not be symmetrized properly, since for example $a_1^\dagger(a_2^\dagger)^3 |\mathrm{vac}\rangle=|1\rangle|3\rangle $ is neither a symmetric nor an antisymmetric state.

*How do we even know what the proper symmetry condition ought to be? Does the wavefunction have to be symmetric under exchange of the phonon mode?

*Does (2.) depend on whether the interacting particles from the start are bosons or fermions, or rather on the type of the mode? What if it is a mix of different kinds or particles?
 A: The Pauli principle is already in full effect - it is enforced at the time that you set your creation and annihilation operators to obey canonical commutation relations of the form
$$
[a_i^\phantom{\dagger},a_j^\dagger]=\delta_{ij},
$$
as opposed to anti-commutation relations of the form $\{a_i^\phantom{\dagger}, a_j^\dagger\} =\delta_{ij}$. For the phonon case you're dealing with the canonical commutation relations are a consequence of the underlying structure rather than an externally-imposed axiom, but the result is the same, as it is the (anti)commutation relations that mark the distinction between fermions and bosons on any second-quantized formalism like the one you're using.
Among other things, this means that your claim that

for example $a_1^\dagger(a_2^\dagger)^3 |\mathrm{vac}\rangle=|1\rangle|3\rangle $ is neither a symmetric nor an antisymmetric state

isn't correct - the state $|1\rangle|3\rangle$ already is fully symmetrized. What matters isn't the wavefunction's symmetry under, in your words,

exchange of the phonon mode,

which doesn't make sense - what you'd care about is exchange of the phonons within each mode or between different modes: i.e. a symmetry operation that takes one of those three photons inside that $|3⟩$ and exchanges it for the second one of those. Or a symmetry operation that takes the photon in the $|1⟩$, puts it in the $|3⟩$, and then takes one of the original $|3⟩$ and puts that inside the same mode that started in $|1⟩$. 
When phrased like that, of course, those symmetry operations don't even make any sense at all, and that is because you're already operating on an automatic second-quantized formalism which, regardless of where it came from, renders such questions completely moot. The exchange symmetry is encoded in the (anti)commutation relations and that's it.
