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.
Yes, the two are isomorphic, there are plenty of ways to see that. I'll begin by sketching a topological argument and then make it more concrete below. If you're not comfortable with algebraic topology you can skip the first part (except the definition of $EG$) up to "why yes there is"
If you're comfortable with algebraic topology, here's one way : note that what you wrote $C_\bullet$ is actually the singular chain complex of the simplicial set $EG$ which is defined in the obvious way, and has a free $G$-action. Moreover, this simplicial set $EG$ is contractible, as it is the nerve of a contractible groupoid (this groupoid is simply the groupoid that has $G$ as objects, and one arrow precisely between any two objects).
It follows that $C_\bullet$ is a projective resolution of the $G$-module $\mathbb Z$, so that what you called $H_\bullet(G)$ is in fact $\mathrm{Tor}^{\mathbb Z[G]}_\bullet(\mathbb Z, \mathbb Z)$
On the other hand, let's have a look at $H_\bullet(BG)$ : $BG$ is a simplicial set which is the nerve of the group $G$. It is therefore a Kan complex, and one can easily compute its homotopy groups to be $G$ for the fundamental group and $0$ for higher homotopy groups.
Therefore if you take a universal covering $\tilde EG\to |BG|$ (the notation is not a coincidence, I'll explain later - note in the meantime that such a universal covering has $\tilde EG/G \cong |BG|$), it has a free $G$-action and is contractible; so that its singular chain complex $C_\bullet(\tilde EG)$ is a projective resolution of $\mathbb Z$ as a $G$-module, therefore to compute $\mathrm{Tor}^{\mathbb Z[G]}_\bullet (\mathbb Z,\mathbb Z)$, one may use this resolution and take $\mathbb Z\otimes_{\mathbb Z[G]}C_\bullet(\tilde EG)$, which, by $\tilde EG/G\cong |BG|$ is nothing but $C_\bullet(|BG|)$, whose homology is $H_\bullet(|BG|)$, and this coincides with simplicial homology, i.e. $H_\bullet(BG)$.
So the two homologies are isomorphic, as they are isomorphic to $\mathrm{Tor}^{\mathbb Z[G]}_\bullet (\mathbb Z,\mathbb Z)$.
Now this is all very abstract for an actually very concrete thing : take the geometric realization of $EG$ as above : it is contractible and has a free action of $G$, the quotient $|EG|/G$ is therefore a space with exactly one nonzero homotopy group, and it's equal to $G$ : if you know a bit about homotopy theory, this tells you that $|EG|/G$ is homotopy equivalent to $|BG|$. This suggests to maybe look at something earlier in our chain of functors: is there actually something relating $EG/G$ and $BG$ ?
Why yes, there is : $EG$ is the nerve of a certain category, $BG$ of another one, can we get a functor between the categories ?
Well send any object $g\in EG$ to the single object $*$ of $BG$ (here I'm seeing them both as categories), and the unique morphism $g\to h$ in $EG$ can be sent to $h^{-1}g \in G$ (I had written $hg^{-1}$ initially, it works too, but it won't work later because of the side you chose for the action - this one will fit nicely with the side of the action). One checks easily that this yields a functor $EG\to BG$ and so a map of simplicial sets.
Moreover, at the level of categories, one can clearly see that this exhibits $BG$ as $EG/G$, so this works for simplicial sets too.
Note that the connection between the nerve of $EG$ and your construction is as follows: an $n$-simplex of said nerve is a composable string of arrows $g_0\to ...\to g_k$, but as there is exactly one arrow between any two objects, this amounts to the list $(g_0,...,g_k)$ of objects (i.e. elements of $G$), and the $i$th boundary map, which corresponds to composition for usual nerves, corresponds to just deleting $g_i$ from the list in our case : so your $C_\bullet$ was indeed the simplicial chain complex of $EG$.
Now what does our functor $EG\to BG$ look like on the level of nerves ? Well it sends $(g_0,...,g_k)$ to $g_0\to...\to g_k$ to $*\overset{g_1^{-1}g_0}\to ... \overset{g_k^{-1}g_{k-1}}\to *$, that is, the map on chain complexes is, very concretely :
$(g_0,...,g_k) \mapsto (g_1^{-1}g_0,...,g_k^{-1}g_{k-1})$
You can now check that this is $G$-equivariant (with the trivial action on the right), and that it is a map of chain complexes (simply because it was a map of simplicial sets to begin with and so it commutes with boundaries ! you can also see it more concretely if you wish : if you remove $g_i$ on the left, then you can multiply $g_{i+1}^{-1}g_i$ and $g_i^{-1}g_{i-1}$ on the right, and you get the correct thing)
Since it is $G$-equivariant, it factors through $(C_\bullet)_G \to \mathbb ZBG_\bullet$, and you can in fact now check that this is an isomorphism (it will come from the fact that $C_\bullet$ is a free $\mathbb Z[G]$-module, and so taking the coinvariants amounts to taking a quotient)
Finally you have to understand that for any $\mathbb ZG$-module $M$, $M_G \cong \mathbb Z\otimes_{\mathbb Z[G]}M$
tldr: Yes, they are isomorphic, and the isomorphism comes from something much earlier in the creation of these chain complexes : it comes from the categories of whose nerve those complexes are the simplicial chain complexes
Best Answer
This follows from the fact that cohomology (with coefficients in a field) is dual to homology. Note first that $H_*(G,\mathbb{F}_p)$ always is the colimit of the $H_*(G_n,\mathbb{F}_p)$ (homology always preserves homotopy colimits). But the dual of a colimit is just the limit of the duals (this is immediate from the universal property of the colimit). So, taking duals of homology, we conclude that $H^*(G,\mathbb{F}_p)$ is the limit of the $H^*(G_n,\mathbb{F}_p)$.
As mentioned in the comments, it is not true with integer coefficients: $\mathbb{Q}$ is the colimit of $G_n=\frac{1}{n!}\mathbb{Z}$ but $H^2(\mathbb{Q},\mathbb{Z})$ is nontrivial while $H^2(G_n,\mathbb{Z})$ is trivial for all $n$. The proof above fails for integer coefficients since cohomology is no longer simply the dual of homology (and indeed, the nontriviality of $H^2(\mathbb{Q},\mathbb{Z})$ comes from the Ext term in the universal coefficient theorem).