Just to sum up, mostly for my own reference, but I thought others might find it useful. (I am new to the site, so please excuse me if this shouldn't be an answer...)
First some preliminary notions:
For a topological space $X$, an $n$-simplex in $X$ is a continuous map $\Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ into $X$. The maps $d^i: \Delta^{n-1} \to \Delta^{n}$, sends $\Delta^{n-1}$ to the face of $\Delta^n$ sitting opposite the $i$th vertex of $\Delta^n$.
An ordered $n$-simplex is a partially ordered set $n_+ = \{ 0 < 1 < \cdots < n \}$. The $n+1$ elements of $n_+$ is called the vertices of $\sigma$. The subsets of $n_+$ are called the faces of $\sigma$. There are morphisms of simplices $d^i: (n-1)_+ \to n_+$ called coface maps, given by $d^i((n-1)_+) = \{ 0 < 1 < \dots < î < \cdots < n \}$ omitting the $i$th vertex of $n_+$.
Then for the two homologies:
The singular (unreduced) chain complex on a space $X$, is the chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(X) \xrightarrow{\partial_n} C_{n-1}(X) \xrightarrow{\partial_{n-1}} \cdots C_1(X) \xrightarrow{\partial_1} C_0(X) \to 0$$
where $C_n(X)$ is the free abelian group $\mathbb{Z}[S_n(X)]$ generated by the set $S_n(X) = \{ \sigma : \Delta^n \to X \}$ of all $n$-simplices in $X$ (i.e. the set of all continuous maps $\Delta^n \to X$). The boundary maps $\partial_n : C_n(X) \to C_{n-1}(X)$ is given by $\partial_n (\sigma) = \sum_{i=0}^{n}(-1)^i \sigma d^i : \Delta^{n-1} \to \Delta^n \to X$.
The $n$th homology group $H_n(X) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ of this complex is the $n$th singular homology group of $X$.
A simplicial complex $S$ is a set $S = \bigcup_{n=0}^{\infty} S_n$ where $S_n = S(n_+)$ being a set of ordered $n$-simplices, such that a face of any simplex in $S$ is itself a simplex in $S$. The simplicial chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(S) \xrightarrow{\partial_n} C_{n-1}(S) \xrightarrow{\partial_{n-1}} \cdots C_1(S) \xrightarrow{\partial_1} C_0(S) \to 0$$
consists of the free abelian groups $C_n(S) = \mathbb{Z}[S_n]$ generated by the $n$-simplices. The boundary map $\partial_n : C_n(S) \to C_{n-1}(S)$ is given by $\partial_n(\sigma) = \sum_{i=0}^n (-1)^i d_i \sigma$ where $d_i = S(d^i) : S_n \to S_{n-1}$ is the face maps $d_i(\sigma) = \sigma \circ d^i$.
The $n$th homology groups of this complex $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ is the $n$th simplicial homology group of $S$.
Lastly we have the realization of $S$, $|S| = \coprod (S_n \times \Delta^n) / \left((d_i \sigma, y) \sim (\sigma, d^iy) \right)$ for all $(\sigma, y) \in S_n \times \Delta^{n-1}$, where $d_i \sigma \times \Delta^{n-1}$ is identified with the $i$'th face of $\sigma \times \Delta^n$.
Then if you want to say something about a specific space $X$, you need to find a simplicial complex $S$, whose realization is homeomorphic to $X$ (i.e. you triangulate $X$ and find the homology groups of the resulting simplicial complex).
NOTE: Feel free to edit any mistakes and clarify where you find it necessary. I'm still not 100% comfortable with it yet..
Consider the simplical set $EG$. It has a free $G$-action and is contractible, so that $C_*(\mathbb Z[EG])$ is a complex of free $\mathbb Z[G]$-modules with a quasi-isomorphism $C_*(\mathbb Z[EG]) \to \mathbb Z[0]$.
$EG$ in degree $n$ is $G^{n+1}$ so this chain complex is $\mathbb Z[G^{n+1}]$ in degree $n$, and the differential is $\sum_i (-1)^i d_i$.
($d_i$ is the one I mentioned in the question, and the $G$-action is coordinate-wise by translation)
The point is that the writing $\mathbb Z[G^{n+1}]$ doesn't make the free-ness apparent, and one way to do so will introduce the $G$-action on $d_0$.
Namely, if $X$ is an arbitrary $G$-set, then $G\times X^{triv}\cong G\times X$ via $(g,x)\mapsto (g,gx)$ (this is obviously a bijection, and it is a simple computation to check that it is $G$-equivariant). In particular, $G^{n+1}\cong G\times (G^n)^{triv}$ via $\varphi: (g,x)\mapsto (g, g^{-1}x)$. One issue with this is that it doesn't respect the simplicial structure, but of course, it shouldn't, and can't ! $EG$ is not $G\times -$ as a $G$-simplicial set. A bigger issue is that the formula for $d_0$ we get is not that good if we do this. But luckily, in this case, we can trivialize the $G^n$ part in several steps.
Indeed, $G^{n+1} = G^{n-1}\times G\times G\cong G^{n-1}\times G\times G^{triv}$, and then you can again remove one $G$ by doing $G\times G\times G^{triv}\cong G\times (G\times G^{triv})^{triv}$, and so on and so forth. If you work this out, you'll see that the isomorphism is given, on $G^{n+1}$, by $G\times (G^n)^{triv}\cong G^{n+1}, (g_0,...,g_n)\mapsto (g_0,g_0g_1,g_0g_1g_2,...., g_0...g_n)$ (you can check a posteriori that this is a $G$-map, and a bijection). Call this $\varphi^{-1}$.
If we compute what $d_i$ looks like with this presentation, we see that for $0\leq i< n$, $\varphi \circ d_i \circ \varphi^{-1} (g_0,...,g_n) =\varphi\circ d_i (g_0, g_0g_1,...,g_0...g_n) = \varphi (g_0,g_0g_1, ...,\widehat{g_0...g_i} ,..., g_0...g_n) = (g_0,...,g_ig_{i+1},...,g_n)$
For $i=n$, $\varphi \circ d_n \circ \varphi^{-1} (g_0,...,g_n)= \varphi\circ d_n(g_0,g_0g_1,...,g_0...g_n) = \varphi(g_0,g_0g_1,...,g_0...g_{n-1}) = (g_0,...,g_{n-1})$.
The important point is to observe that for $i=0$, this does not send $\{e\}\times G^n$ to $\{e\}\times G^{n-1}$
So if we write the $n$-term of our chain complex as $\mathbb Z[G]\otimes_\mathbb Z\mathbb Z[(G^n)^{triv}]$ to witness that it is indeed a free $G$-abelian group, and compute the differential on $\hom_G(\mathbb Z[G]\otimes_\mathbb Z[G^n],A)$, we see that if we view elements of this as set maps $G^n\to A$, then for all differentials except for $d_0$, we have nothing to do, but precomposition by $d_0$ makes us go out of $G^n$ so there is something to say.
Precomposition by $d_0$ looks like this : if $f: G^n\to A$, then we can view $f$ as a $G$-map $\tilde f: G\times G^n\to A$ by sending $(g_0,...,g_n)$ to $g_0\cdot f(g_1,..,g_n)$.
So now, the map $G^{n+1}\to A$ associated to $\tilde f\circ d_0$ is defined by $\tilde f\circ d_0 (e,...,g_{n+1}) $, and this is $\tilde f(g_1,..., g_{n+1}) = g_1\tilde f(e,g_2,...,g_{n+1}) =g_1 f(g_2,...,g_{n+1})$.
So this gives us the desired $d_0$, where the other $d_i$'s preserve $\{e\}\times G^n$, so there is nothing to say.
Best Answer
I believe I have settled this issue for myself, and the key ingredient is indeed the Dold-Kan correspondence. However, I find a synthesis of these two expositions to be much more comprehensible than anything nlab has to offer: (1), (2) They also both refer to the same book reference, namely this one, which is also helpful in understanding the result.
The basic idea is that any chain complex corresponds to a simplicial abelian group, and any simplicial abelian group $A$ has a corresponding chain complex $A_*$. The image of the degenerate simplices under this correspondence forms a subcomplex $DA_*$, and there is another subcomplex $NA_*$ which turns out to have the property that $NA_*\oplus DA_*\cong A_*$. In other words, $A_*/DA_*\cong NA_*$. The Dold-Kan correspondence theorem itself says that $A\to NA_*$ gives an equivalence of categories between chain complexes and simplicial abelian groups, which then gives a way to "normalize" certain chain complexes, including the chain complex defining the simplicial homology of a topological space.
In order to apply this to my situation, which involves cohomology rather than homology, it is necessary to "dualize" this result. The key insight, I think, is that actually any simplicial set $X$ can be "upgraded" into an abelian simplicial group $\mathbb Z (X)$, whose $n$-simplices $\mathbb Z(X)[n]$ are $\mathbb Z (X[n])$, which consists of formal sums $\sum_{x\in X}k_x x$ such that $k_x\in \mathbb Z$ and only finitely many of them are nonzero. Then any cochain complex $C^n(X,A)$ defined as $\lbrace f:X[n]\to A \rbrace$ can instead be thought of as $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$, and if $C_0^n(X,A)$ is the functions which are zero on any nondegenerate $n$-simplex, then $C^n(X,A)/C^0_n(X,A)\cong Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$ and the Dold-Kan correspondence gives a chain equivalence between $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$ and $ Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$, which makes the natural projection $C^n(X,A)\to C^n(X,A)/C^0_n(X,A)$ into a chain equivalence also.
After this, I found that it does appear necessary, or at least convenient, to go through $\mathcal E G$, the simplicial set with $\mathcal E G[n]=G^{n+1}$ and $d_i$, $s_i$ given by deletion and repetition of the $i$th element respectively, and whose geometric realization is $EG$. Even though this (or the bar resolution) may be viewed as a simplicial group, $G$ may not be abelian, so that isn't particularly helpful. It's better to think of it as a simplicial set and use a slightly modified version of the result above (essentially, the same formulation can be done for subcomplexes of $C^n(X,A)$ to "normalize" them). Then (in broad strokes) if $C^n(G,A)=\lbrace f:G^{n+1}\to A\text{ such that }gf=fg\text{ for all }g\in G \rbrace$, there is an injective map $C^n(G,A)/C^n_0(G,A)\to C^n_\text{cell}(EG,A)$, and an injective map $C^n_\text{cell}(BG,A)\to C^n_\text{cell}(EG,A)$; if $g$ acts trivially on $A$ then both of which have the same subcomplex as their image. The isomorphism follows from that, and if $G$ doesn't act trivially on $A$ then that first injective map can still be used to get an isomorphism to the cohomology with local coefficients instead.