For instance, section 4 of arxiv.org/abs/1708.07192 in another word, take $\hbar$ to 0. Of course this doesn't say anything about the state, merely the hamiltonian. But the states (at least the product states) can be obtained similarly.

Actually I don't get mixed state at all! As I mentioned in the question, there is no "thermodynamics" after taking the classical limit. For instance, I take a ferromagnetic state of a 1D spin chain, the classical limit is still a ferromagnetic "configuration" in my classical theory (say ferromagnetic spin wave theory). (of course it is almost impossible to describe a highly entangled quantum state à la this). In the classical field theory, it is just an initial value problem, which does NOT have a thermodynamic entropy for instance.

For instance, extended Bose Hubbard model is non-integrable, however, in the phase diagram, there exists a $c=1/2$ CFT, i.e. "integrable" as a CFT, as explained explicitly in this article, arxiv.org/abs/1812.03489

I don't quite understand your setup. Following your example, the Hamiltonian is completely decoupled from site to site, as you mentioned. Therefore, for any $n$, one just have to calculate spectrum from each site (the chemical potential and Hubbard interaction commute) and add them up . There is no need to discuss about integrability at all. If you want that, it is exactly solvable without using any technique like Bethe ansatz.