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I agree with your argument, but if by $C_n$ in your final formula you mean the cyclic group of order $n$, then I don't see exactly where that is coming from. Naively I would think that for the corners the group should be $A_4$, and for the cubies with 3 stickers it should be $S_3$. More generally I'd expect for a $d$-dimensional cube that the corners correspond to $A_d$ and all other types of pieces correspond to $S_k$ for $k=2,\ldots,d-1$. Could you explain how you get that (or if I'm misunderstanding your claim)?
Maybe I'm missing something obvious, but it seems to me that the standard proof for ordinary Rubik's cubes works for $3^n$ cubes for any $n$ to give the cube group as a subgroup of a direct product of wreath products, with each factor corresponding to acting on cubies with $k$ stickers for $k=2,3,\ldots,n$. Is this what you want, or are you looking for something else?
@FedericoCarta I'll also note that questions should be tailored to the community. The original questions you posted were decent for physicists, though they were too broad together and should have been split up. However, if you're going to ask the same question on a site for research mathematicians, you need to be more tailored to that community. CFTs are of interest to mathematicians as well as physicists, but the literature in the two disciplines is quite different. I'm not sure which one you wanted to learn, but if it's the physics version this would probably not be the right place to ask.
@FedericoCarta Yes, different sites have different standards, and cross-posting isn't always bad. However, cross-posting immediately is frowned upon, whether or not the question was closed. This MSO post details the proper protocol for if you think you posted on the wrong site. Immediately cross-posting a closed question to a different site with no indication that it is a cross-post can come across as trying to game the system (I don't thin this was your intention). At minimum you should try to improve your post there first before reposting it here.