In this answer, it is pointed out that in the context of Lie algebra cohomology, the Lie algebra $\mathfrak g$ acts trivially on $\mathbb C$, i.e. $$a\cdot c=0$$ for all $a\in\mathfrak g,c\in\mathbb C$. This makes sense to me.
However, if that is true, I do not understand why we need the first term on the r.h.s. of the formula for the exterior derivative:
$$ \mathrm{d}f(x_1,\dots,x_{n+1})= \sum_i(-1)^{i+1}x_i f(x_1,\dots,\hat{x}_i,\dots,x_{n+1})+\dots $$
If the Lie algebra elements $x_i$ act trivially on $\mathbb C$, and $f:\bigoplus_n\mathfrak g\to\mathbb C$, then shouldn't $x_i f(x_1,\dots,\hat{x}_i,\dots,x_n)$ vanish?
To compute the Lie algebra cohomology $H^i(\mathfrak{g},M)$, where $M$ is a $\mathfrak{g}$-module, the Chevalley complex is formed by placing in degree $n$ all linear maps $f:\Lambda^n\mathfrak{g}\to M$, and using the differential you are referring to above. If $M$ is not a trivial $\mathfrak{g}$-module, then the actions of $x_i$ on $f(x_1\wedge\dots\wedge\hat{x_i}\wedge\dots\wedge x_n)$ will in general not be trivial.
On the other hand, if $M$ is a trivial $\mathfrak{g}$-module, (e.g. if $M=\mathbb{C}$ with the trivial action), then indeed $x_if(x_1\wedge\dots\wedge\hat{x_i}\wedge\dots\wedge x_n)=0$.