In tetrad formalisms you don't want the Christoffel symbols $\Gamma_{abc}$, you want the connection 1-forms. For a given basis 1-form $e^a$, $d(e^a)$ is a 2-form. Now it must be that $$d(e^a) = \omega^a{}_b \wedge e_b.$$
for a matrix of 1-forms $\omega^a{}_b$. Actually, it's usually better to think of $\omega^a{}_b$ as a matrix-valued one-form. In fact, it takes values only in $\mathfrak{so}(1,3)$, the Lie algebra of the Lorentz group. More concretely, that means that $\omega^a{}_b$ is antisymmetric with respect to the Minkowski metric.
Anyway, in this formalism, the Riemann curvature is an $\mathfrak{so}(1,3)$-valued 2-form given by $$R^a{}_b = d\omega^a{}_b + \omega^a{}_c \wedge \omega^c{}_b.$$
The statement of the Bianchi identity is $$dR^a{}_b + \omega^a{}_c \wedge R^c{}_b + R^a{}_c \wedge \omega^c{}_b = 0.$$
You will find this material in Sections 14.5 and 14.6 of Gravitation by Misner, Thorne, and Wheeler.
The above does not connect to the Newman-Penrose formalism since it doesn't mention spinors. To obtain the NP formalism, you must use
dyads, the spinor lifts of null tetrads. In this formalism the connection 1-form is a $2\times 2$ antisymmetric complex matrix, so there are $3\cdot 4=12$ components, each of which can be a given its own Greek letter. The Newman-Penrose equations are now obtained by writing out in dyad components the equations above, representing tetrad derivatives by $D, \Delta, \delta, \overline{\delta}$.
I have never seen this done in detail as I imagine the calculations are horribly tedious and boring.