The covariant derivative is initally defined on vector fields and then it is extended to all kinds of tensor fields by assuming that (a) this action is linear, (b) it works as a derivative with respect to the tensor product:

$$\nabla_X (u \otimes v) = (\nabla_X u) \otimes v + u \otimes \nabla_Xv \:,$$

(c) it acts as a standard vector field when acting on scalar fields: $\nabla_X f = X(f)$, and (d) it commutes with contractions:

$$ \nabla_X \langle Y, \omega \rangle = \langle \nabla_X Y, \omega \rangle + \langle Y, \nabla_X\omega \rangle\:.$$

With this definition, it is difficult to say if this question makes sense. Referring to a single connection coefficients, i.e., for fixed values of $\alpha,\beta,\gamma$:

$$\Gamma^\alpha_{\beta\gamma} = \langle \nabla_\beta \partial_{x^\gamma}, dx^\alpha\rangle $$

it makes sense, since it is a scalar field and thus the covariant derivative $\nabla_X$ coincides with the action of the vector field $X$ on $\Gamma^\alpha_{\beta\gamma}$.

However, I do not think it was the intention of the OP. I guess that, instead, the OP thinks of the whole set of "components" $\Gamma^\alpha_{\beta\gamma}$ for all values of the indices. Unfortunately this set does not define a tensor so we cannot apply (a),(b),(c),(d) to this set of components. The only thing one can say is that the differences of connection coefficients referred to two different connections define a tensor field (the proof is evident) of components,

$$T^\alpha_{\beta \gamma}= \Gamma^\alpha_{\beta \gamma} - \overline{\Gamma}^\alpha_{\beta \gamma}\:.$$

Consequently, the covariant derivative (w.r. to a third affine connection) of this difference is well defined.

What I am saying is that I do not think it make sense to assume that the Riemann tensor is constructed out of something like covariant derivatives of connection coefficients...