From the adjunction $U:\mathsf{Grp}\leftrightarrows\mathsf{Set}:F$ between the free group functor and the forgetful functor, one gets a monad $F\circ U:\mathsf{Grp}\to\mathsf{Grp}$, which we can write just $F$. Given a group $G$, there is an associated (augmented) simplical group $F^\sharp G$, the bar construction for the monad, which looks like the one you wrote.

If $M$ is a $G$-module, them $M$ is a $F^nG$-module for all $n$, by pulling back along the composition of maps going from $F^nG$ to $G$ in the simplical group. It makes sense to talk about derivations from the $F^nG$ to $M$, and in fact we get a cosimplicial abelian group $\operatorname{Der}(F^\sharp G,M)$. From it we can construct as usual a cochain complex (whose differentials are alternating sums of transposed face maps in the simplicial group $F^\sharp G$)

(I have not completely check the details but) the cohomology of the complex $\operatorname{Der}(F^\sharp G,M)$ should be isomorphic to the cohomology of $G$ with values in $G$ up to a shift and a twist: explicitly, $$H^i(\operatorname{Der}(F^\sharp G,M))\cong\begin{cases}Der(G,M), &\text{if $i=0$;}\\H^{i+1}(G,M), &\text{if $i>0$.}\end{cases}$$ This is the usual group cohomology of $G$, but we dropped the $0$th group, and instead of having derivations modulo the inner ones, we have just derivations. Most of the work needed to check this is of the general nonsense kind.

If you apply a more interesting functor $\mathsf{Grp}\to\mathsf{Ab}$ to $F^\sharp G$, you'll get other things, of course.

The following does not answer your question, but shows that the object you have in mind does give useful information (notice I did not drop all but the outer evaluation maps but kept the whole simplicial group).

From the adjunction $U:\mathsf{Grp}\leftrightarrows\mathsf{Set}:F$ between the free group functor and the forgetful functor, one gets a monad $F\circ U:\mathsf{Grp}\to\mathsf{Grp}$, which we can write just $F$. Given a group $G$, there is an associated (augmented) simplical group $F^\sharp G$, the bar construction for the monad, which looks like the one you wrote.

If $M$ is a $G$-module, them $M$ is a $F^nG$-module for all $n$, by pulling back along the composition of maps going from $F^nG$ to $G$ in the simplical group. It makes sense to talk about derivations from the $F^nG$ to $M$, and in fact we get a cosimplicial abelian group $\operatorname{Der}(F^\sharp G,M)$. From it we can construct as usual a cochain complex (whose differentials are alternating sums of transposed face maps in the simplicial group $F^\sharp G$)

(I have not completely check the details but) the cohomology of the complex $\operatorname{Der}(F^\sharp G,M)$ should be isomorphic to the cohomology of $G$ with values in $G$ up to a shift and a twist: explicitly, $$H^i(\operatorname{Der}(F^\sharp G,M))\cong\begin{cases}Der(G,M), &\text{if $i=0$;}\\H^{i+1}(G,M), &\text{if $i>0$.}\end{cases}$$ This is the usual group cohomology of $G$, but we dropped the $0$th group, and instead of having derivations modulo the inner ones, we have just derivations. Most of the work needed to check this is of the general nonsense kind.

If you apply a more interesting functor $\mathsf{Grp}\to\mathsf{Ab}$ to $F^\sharp G$, you'll get other things, of course.

All the above works because free groups play particularly nice as the domain of derivations: they behave like projectives. AAndré-Quillen cohomology resolves general commutative rings with polynomial rings, because it wants to find derivatives of the functor of derivations (or of Kähler forms, in the homology case) and for these polynomial rings are "projective")