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The Pauli Hamiltonian is $$H= \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi $$
For the 1-component Schrodinger equation in absence of applied field we have the Lagrangian $$\mathcal{L} = i\hbar\psi^*\dot \psi - \frac{\hbar^2}{2m}\Big( \boldsymbol{\nabla}\psi^*\cdot\boldsymbol{\nabla}\psi \Big)- \psi^*V\psi $$ What is the 2-component analog of this which corresponds to the Pauli Hamiltonian?
Well, it is a straightforward transcription, if you appreciate the language.
For the 1-component (classical) Schroedinger field in absence of applied external EM fields, the Lagrangian density is actually $$\mathcal{L} = i\hbar\psi^*\dot \psi - \frac{1}{2m}\Big( \psi^*\boldsymbol{p}^2\psi \Big)- \psi^*V\psi ~~;$$ (I have integrated your expression, corrected, by parts and supplanted ${\boldsymbol p}=-i\hbar {\boldsymbol \nabla}$.)
The usual degenerate canonical momentum for ψ is but $\Pi_\psi=i\hbar \psi^* $, which eliminates the time derivative term in the Legendre transform to the hamiltonian, since $\psi^* \Pi_\psi-i\hbar \psi^* \partial_t\psi=0$.
The ensuing field Hamiltonian density is then also trivial, $$ {\cal H}= \psi^* \Big( \frac{1}{2m} \boldsymbol{p}^2 +V\Big) \psi, $$ netting you the standard Schroedinger hamiltonian.
To generalize to Pauli's, promote the wave-fields/functions to two-spinors, insert the requisite vector potentials (in the non-quantized classical potential part), and you have $$\mathcal{L} = i\hbar\psi^\dagger\cdot \dot \psi - \psi^\dagger \cdot \Big( \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi \Big) \psi.$$ Recall the Pauli matrices are Hermitian. The action is the space & time integral of this.
Again, variation w.r.t. the classical two-spinor fields/variables (like the qs and ps of classical mechanics) yields the corresponding Schroedinger/Pauli equation.
