I think you're right, in the general case we should consider temperatures, but that is a matter of definition and I think cohesive energy is defined for zero temperature. In this particular case, I will try to answer your questions for $T=0$.
We will do this by neglecting the kinetic energy of the nuclei (phonons). What essentially happens is at $T=0$ we go to the ground state of the system, and the ground state should not have any phonons. The general process goes like this - calculate the energy of an electron in an Isolated atom and multiply it by $N$, which is the number of atoms. Then you compare this with the energy of the ground state of the electron Hamiltonian in the lattice. The ground state (neglecting interactions) will look like the electrons occupying the $N$ lowest energy states. So the energy of the lattice will be the sum of the energies of all these states, which will generally be lower than the energy of $N$ isolated atoms (otherwise we wouldn't have solids!). This difference will give us the cohesive energy.
Now let's try to do the tight binding problem. \begin{equation} H = -E_0\sum_n |n\rangle\langle n|-V\sum_n (|n\rangle\langle n+1| +|n+1\rangle\langle n|) \end{equation} This is an approximation to a system of non-interacting electrons in a 1D lattice. 976
$-E_0$ is like the energy of an electron in an isolated atom and when you bring these atoms together the $V$ terms lets the electron `hop' between neighboring sites. But this changes the system drastically. Like you mentioned, we get the dispersion \begin{equation} E(k) = -E_0 - 2V \cos (ka) \end{equation} where $a$ is the lattice constant. This gives us the density of states $$ g(E) \sim \frac{1}{\sqrt{1-(\frac{E+E_0}{2V})^2}} $$ An understandable assumption here would be that each atom donates one electron to the system, so we have a total of $N$ electrons and $N$ atoms. But this means that the whole band is filled, and if you integrate to calculate the energy of this system, you will get exactly $-NE_0$, which is equal to the energy of $N$ isolated atoms. So the cohesive energy is zero, which means such a crystal will not form due to electron delocalization. Note: if you have learnt some Molecular Orbital theory, this is exactly why the molecule $He_2$ does not exist!
The idea of Fermi temperature is not very useful here since there are no electronic excitations possible but if one were to define such a quantity , it would be $T_f = \frac{2V}{k_B}$ since it should not depend on $E_0$, which simply shifts the energy of all states. So we choose it such that the lowest energy state is at $T=0$.
Why do we have metals, then? What did we miss? The thing we missed is that metals generally have free orbitals at energies close enough to the energy of the orbital (don't forget spin degeneracy!) the electron is in ($-E_0$), so we have additional bands where the electrons go in and reduce their energy.
One final comment on temperature - I am not sure about the exact definition of cohesive energy, but if you wanted to use cohesive energy to see if a solid forms at any finite temperature, you have to take entropy into account. The state that minimises the <PERSON>/Gibbs free energy will be observed.
Some nice textbooks for this would be The Oxford Solid State Basics by <PERSON> and the classic textbook by <PERSON> and <PERSON>. 28