Constructing Dirac's Lagrangian without assuming hermitian/antihermitian $\gamma$ matrices The Dirac $\gamma$ matrices follow the Clifford algebra:
\begin{equation*}
\{ \gamma^\mu,\gamma^\nu\} = 2\eta^{\mu \nu}
\end{equation*}
The traditional construction of the Dirac Lagrangian involve choosing a basis of the $\gamma$ matrices where $\gamma^\mu$ is either Hermitian or antihermitian depending on the metric, and proceeding to define the Dirac adjoint $\bar{\psi} = \psi^\dagger \gamma^0$. The textbooks then proceed to show that $\bar{\psi}\psi$ is a Lorentz scalar, etc. and construct the Dirac Lagrangian with the 'kinetic' term and the mass term.
Is it possible to construct the Dirac Lagrangian WITHOUT using the hermiticity of the $\gamma$ matrices? As far as I can see, the $S[\Lambda]$ matrix defined by the non-hermitian $\gamma$ matrices is a valid representation of the Lorentz group. So why do the textbooks resort to the hermiticity of the  $\gamma$ matrices to construct the Lagrangian?
\begin{equation*}
\mathcal{L} = i\bar{\psi}\displaystyle{\not}\partial\psi -m \bar{\psi}\psi
\end{equation*}
If we construct the Dirac Lagrangian without assuming the hermtiticty of these matrices, how do we now define the Dirac adjoint?
 A: The answer is no. The gamma matrices are defined by the Clifford algebra
$$\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}$$
Let's say we are working in the signature with $\eta^{\mu\nu}=\text{diag}(-1,+1,+1,+1)$. We have
$$(\gamma_0)^2=-1,\quad(\gamma_i)^2=1$$
These equations say that $\gamma_0$ has eigenvalues $\pm i$ (hence anti-Hermitian), and the $\gamma_i$'s have eigenvalues $\pm1$ (hence Hermitian).
I see what you are saying about the generators $\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}]$ being the important object for defining the representation, and that there may be room for choosing $\gamma^{\mu}$ which leaves the commutator invariant. But the commutator works as a generator of the Lorentz group as a consequence of the Clifford algebra.
Edit
What I wrote above is incorrect. In fact, one can find solutions to the Clifford algebra which are not Hermitian. Here's a set of solutions I found of the $2d$ Clifford algebra
$$\gamma_0=i\begin{pmatrix}1&A\\0&-1\end{pmatrix},\quad\gamma_1=\begin{pmatrix}-1&0\\2/A&1\end{pmatrix}$$
You can probably play with the $4d$ case to find examples but it escapes me at the moment. I think a better answer as to why we don't use gamma matrices like this is they result in generators which are neither Hermitian nor anti-Hermitian. Typically it is much harder to make a theory unitary if the representations we use of whatever symmetry group are not unitary representations. Now, the Lorentz representations we use are not unitary, but the generators are either Hermitian or anti-Hermitian (rotation and boost generators respectively), and there is a procedure colloquially known as Wick-Rotation which enables us to define a unitary theory in this case.
