Covariant derivative of connection coefficients

The formalism is explained very well in Landau-Lifshitz, Vol. II, par. 92 (properties of the curvature tensor). The Riemann curvature tensor can be called the covariant exterior derivative of the connection. The exterior derivative is a generalisation of the gradient and curl operators.

You might also consider looking at the geometry in differential forms language. The connection is seen as a 1-form (to be integrated along a line, the corresponding index is supressed), resulting in a (2-index) transformation matrix.* The Riemann curvature tensor is seen as a 2-form (to be integrated over a surface), again with values in a (2-index) transformation matrix. By doing so, you see Stokes' theorem appear, since integrating the connection (1-form) along a closed lines yields the same result as integrating the Riemann curvature (2-form) over the enclosed surface. That's why the Riemann curvature (2-form) needs to be the covariant exterior derivative of the connection (1-form).

Literature: Nakahara, Geometry, Topology and physics, chap. 5.4 and 7.

*Precisely: a Lie-algebra valued 1-form.

Your question:

Is there a meaningful way to define the covariant derivative of the connection coefficients...?

has a very simple answer: NO

It does not make sense (and it would be a very bad practice) to overload the operator "covariant derivative" and force it to somehow work on objects that are not tensors or scalars.

$\delta\Gamma$ (the variation of $\Gamma$ in an action ) is however a tensor with 3 indices, so you will find expressions like $$\delta\Gamma^a_{bc}$$ for the a,b,c component of this tensor and $$\delta\Gamma^a_{bc;d}$$ the a,b,c,d component of its covariant derivative, and even $$\delta R_{ab}=\delta\Gamma^l_{ab:l}-\delta\Gamma^l_{al:b}$$ for the variation of the Ricci tensor.

All this can be found in MTW page 492 and page 500

Pleasse note that: $$\delta\Gamma^a_{bc;d}$$ does not mean $$\delta (\Gamma^a_{bc;d})$$ which is meaningless, but rather $$(\delta\Gamma)^a_{bc;d}$$ since $\delta\Gamma$ is the tensor being differentiated