When in $T rightarrow infty$ the eigenstates that contribute the most are the ones that lie in the middle of the spectrum?

Is the following intuition correct?

Consider a 1D Ising model of length $N$ in a heat bath with inverse temperature $beta$ (i.e. a canonical ensemble). The Boltzmann factor is given by
$$
p(E_i) = frac{g_i e^{-beta E_i}}{sum_i g_i e^{-beta E_i}},
$$

where the factor of $g_i$ accounts for the multiplicity of macrostate with energy $E_i$ (i.e. its degeneracy).

To further study $p(E_i)$, one can diagonalise the Hamiltonian with an eigenbasis given by the tensor product of the eigenvectors of the $sigma_z$ operator ${|uparrow rangle, |downarrow rangle }$, i.e. with
$$
{underbrace{|uparrow rangleotimes|uparrow rangle otimescdots otimes|uparrow rangle}_text{$N$ times}, |downarrow rangleotimes|uparrow rangle otimescdots otimes|uparrow rangle, cdots, |downarrow rangleotimes|downarrow rangle otimescdots otimes|downarrow rangle }.
$$
It is easy to see, that for a given total magnetisation $M_T = langle S^z_T rangle equiv langle sum_i S^z_i rangle$, when $M_T=0$ the multiplicity of this macrostate is the largest, given by $g={N choose N/2}$ microstates.

Then, $e^{-beta E_i}$ in the limit $lim_{βrightarrow 0}e^{-beta E_i}=1$ and the Boltzmann factor is given by
$$
p(E_i, beta =0)= frac{g_i}{sum_i g_i}.
$$
Thus in the thermodynamic limit, the state with the largest multiplicity $g_i$ dominates i.e. we have
$$
p(E_i,beta=0) sim left{begin{aligned} 1, quad M_T=0 0, quad M_Tneq0 end{aligned}right.
$$
In this particular example, the statistics are dominated by the eigenstates that lie in the middle of the spectrum but whereas this is true for this example, in general, it might not be true in other cases. As @Jahan Claes pointed out, it depends on the shape of the density of states as shown above.