= Solution
With no pairing, the <Hamiltonian> is diagonal in the <fermionic occupation numbers>. The <Pauli exclusion principle> allows each <momentum> mode to be either empty or occupied once. Without a fixed <fermion number> constraint, occupying a mode lowers the <energy> precisely when its single-particle <energy> is negative. Therefore, with $|0\rangle$ the <Fock vacuum>, a <ground state> is
$$
\boxed{|\mathrm{GS}\rangle=\prod_{p:\,\epsilon(p)<0}c_p^\dagger|0\rangle,\qquad E_{\mathrm{GS}}=\sum_{p:\,\epsilon(p)<0}\epsilon(p).}
$$
Choose a fixed ordering of the <fermionic creation operators>; changing it alters only the overall <phase>. Any zero-energy modes may be occupied or empty without changing the <ground-state energy>.
To identify the occupied modes, complete the square:
$$
\frac{\epsilon(p)}\alpha=\left(p^2-\frac\beta2\right)^2+\mu-\frac{\beta^2}{4}.
$$
Here $\beta$ is the dispersion parameter, not <inverse temperature>. If $\beta^2<4\mu$, every mode has positive <energy> and the unique <ground state> is the <Fock vacuum>, with $E_{\mathrm{GS}}=0$. If $\beta^2=4\mu$, there are still no negative-energy modes, but $p=\pm\sqrt{\beta/2}$ have zero <energy> if they lie on the allowed <momentum> grid. Each such mode doubles the <ground-state degeneracy>.
For $\beta^2>4\mu$, define
$$
p_-^2=\frac{\beta-\sqrt{\beta^2-4\mu}}2,\qquad
p_+^2=\frac{\beta+\sqrt{\beta^2-4\mu}}2.
$$
Both are positive, and the occupied <Fermi sea> consists of the two intervals $p_-<|p|<p_+$. With $p=2\pi n/L$, the requested finite-size <ground-state energy> is explicitly
$$
\boxed{E_{\mathrm{GS}}=2\alpha\!\sum_{\substack{n\in\mathbb Z_{>0}\\ Lp_-/(2\pi)<n<Lp_+/(2\pi)}}\left[\left(\frac{2\pi n}L\right)^4-\beta\left(\frac{2\pi n}L\right)^2+\mu\right].}
$$
The factor two counts opposite <momenta>. Endpoints are excluded from this <sum> because they have zero <energy>; when an endpoint is allowed, its optional occupation adds <ground-state degeneracy>. A narrow negative-energy interval can contain no allowed finite-size <momenta>, in which case the <sum> is empty. The zero-momentum mode is always empty since $\epsilon(0)=\alpha\mu>0$.
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