Both (Lie) group and Lie algebra cohomology are essentially part of a more general procedure. Namely, we take an abelian category $C$ with enough projectives or enough injectives, take a (say, left) exact functor $F$ from $C$ to abelian groups (or modules over a commutative ring) and compute the (right) derived functors of $F$ using projective or injective resolutions.
For example, if $mathfrak{g}$ is a Lie algerba over a field $k$, we can take the category of $U(mathfrak{g})$-modules as $C$ and $$Mmapstomathrm{Hom}_{mathfrak{g}}(k,M)$$
as $F$ (here $k$ is a trivial $mathfrak{g}$-module). Notice that this takes $M$ to the set of all elements annihilated by any element of $mathfrak{g}$; this is not a $mathfrak{g}$-module, only a $k$-module, so the target category is the category of $k$-vector spaces.
In the category of $U(mathfrak{g})$-modules there are enough projectives and enough injectives, so in principle to compute the Exts from $k$ to $M$ we can use either an injective resolution of $M$ or a projective resolution of $k$ as $U(mathfrak{g})$-modules.
I've never seen anyone considering injective $U(mathfrak{g})$-modules, probably because they are quite messy. So most of the time people go for the second option and construct a projective resolution of $k$.
One of the ways to choose such a resolution is the "standard" resolution with
$$C_q=U(mathfrak{g}otimesLambda^q(mathfrak{g})$$ but any other resolution would do, e.g. the bar-resolution (which is "larger" then Chevalley-Eilenberg, but more general, it exists for arbitrary augmented algebras). Applying $mathrm{Hom}_{mathfrak{g}}(bullet,M)$ to the standard resolution we get the Chevalley-Eilenberg complex.
All the above holds for group cohomology as well. We have to replace $U(mathfrak{g})$ by the group ring (or algebra) of a group $G$. The only difference is that there is no analogue of the Chevalley-Eilenberg complex, so one has to use the bar resolution. Probably, Van Est cohomology of a topological group can also be described in this way.
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