I came across with the following questions in reading Altland and Simons page 186

Is $$\alpha=l$$ the same orbital quantum number related to angular momentum $$\hat{L}\to l(l+1)$$?

They talk about that fermions couple to a magnetic field by their orbital momentum. Why did not they include that orbital momentum coupling? Are they talking about $$\sim \vec{L}\cdot\vec{B}$$ ?

They write down the hamiltonian $$H_0=\sum_{\alpha, \sigma}a^\dagger_{\alpha\sigma}\epsilon_\alpha a_{\alpha\sigma}$$

What is the expression for $$\epsilon_\alpha$$ ?

$$H_0$$ is the "non magnetic" Hamiltonian operator, which means the part which is not coupled to the magnetic field. In the purpose of the exercise, you can think the quantum number $$\alpha$$ as the set of momentum-spin $$k,\sigma$$. For instance if you have an electron system in a cubic lattice with hopping between nearest neighbours sites, you have $$\varepsilon_{k\sigma} = -2t\sum_{\mu} \cos{(\vec{k} \cdot \hat{e}_{\mu})}$$ (where $$\vec{k}$$ is the momentum and $$\hat{e}_{\mu}$$, $$\mu=x,y,z$$ is the versor of an axis). Anyway the explicit form is not necessary to solve the exercise.
Your doubt is justified: if you want to give a full quantum treatment of the electron gas coupled to a magnetic field, you not only have to take into account the coupling with the spin $$-\mu_B\vec{S}\cdot \vec{B}$$, but also the coupling to the orbital angular momentum $$-\mu_B\vec{L}\cdot \vec{B}$$, which indeed affects $$H_0$$.
However one can prove that the magnetic susceptibility can be written as $$\chi = \chi_L + \chi_S$$, where $$\chi_S$$ is what you get if you consider the spin coupling only, while $$\chi_L$$ is what you get if you consider the orbital coupling only. A reference is Landau Vol. 5 (which is really precise, but a little tough to understand). In conclusion, the spirit of the exercise is to compute only one of these two contributions, in particular $$\chi_S$$.
It turns out that $$\chi_S$$ is positive (paramagnetic term), while $$\chi_L$$ is negative (diamagnetic term), and that $$\chi_L = -\chi_S/3$$ if you assume no underlying lattice structure. In conclusion the free electron gas is paramagnetic, and the effect of the term you are neglecting is to reduce the paramagnetic response by a factor $$2/3$$.