Is there a way to define a Lie derivative of a connection?

Of course, yes. Lie derivative is defined for any geometric object (= when it is defined what happends when we change a coordinate system): take the flow $\phi_t$ of the vector field, consider the pullback $\phi_t^*\Gamma$ of your geometric object $\Gamma$ and define Lie derivative as the $\tfrac{d}{dt}$-derviative at $t=0$ of $\phi_t^*\Gamma$.

For affine connection its Lie derivative is an (1,2)-tensor field. It is because derivative is more or less the same as difference and difference of two connections is a (1,2)-tensor field.

Of course if the Lie derivative of a connection is zero then the connection is presered by the flow so you vector field is a symmetry of the connection.

Of course there exists an algebraic formula for the Lie derivative in terms of the components of an object in the coordinates ($\sim$ Christoffel symbols, for example), components of the vector field, and their first derivatives. I do not remember this formula by hart but Maples DifferentialGeometry package knows it. It is a sum of two terms, the first part is the usual formula of the Lie derivative for a (1,2)-tensorfield and the other part is something like the vector field plugged in the curvature.

I was expecting the Lie derivative of a connection to be another connection but Vladimir's answer about the difference of two connections being a tensor makes a lot of sense. I would say a coordinate free algebraic formula for the Lie derivative of a connection is:

$\left(\mathcal{L}_{X}\nabla \right)_{Y}Z:=\mathcal{L}_{X}\left( \nabla_{Y}Z\right)-\nabla_{\mathcal{L}_{X}Y}Z-\nabla_{Y}\mathcal{L}_{X}Z$.

This is just the formula for the Lie derivative of a tensor and in this case it does give a $C^{\infty}(M)$-linear object in the variables $Y$ and $Z$. Since this is a tensor field probably a better notation would be

$\left(\mathcal{L}_{X}\nabla\right)\left( Y,Z \right)$.