If there is nothing satisfactory, I could also just go for "an approximation"...

It would be nice if there was some conomological stuff to control the twistedness of such object. I am an algebraic topologist, so no expert here, but some substitute of simply connectedness on $G$ could be a nice hypothesis to guarantee the triviality and make OP's formula true.

Thanks Joe! I was by phone and I meant to write 'just a comment', then it became too long and I didn't take time to readjust the format. Much nicer now!

Isn't the $\infty$ category $M_{\infty}$ "just" $\mathcal{N}(M)[\mathcal{W}^{-1}]$?

You are right: checking a given thing is unit is easy, finding them all it's hard

As a side comment, the units are not easy to find/check: see this question. Also, you could try a simplified search by multiplying for the cyclotomic units which are quite explicit: planetmath.org/…. One could hope to find a bound on which cyclotomic units could possibly contribute (maybe some coefficient diverges when you multiply many times?). But then there is the problem of non-cyclotomic units, which implies a finite search from any candidate polynomial.

Nicest question of the year!! Isn't there a badge for that? i think I'll do this question in all my topology-based dissemination :)

Ok!!! I changed it accordingly. For the moment I neglected the difficulty of the bias that i mention at the end, but it should be a "naive" guess :) the formula is not consistent with yours, but now they are much closer!

Can you permute the digits of a single number? For example, is $12340156789$ admitted? In that case, there are many more permutations than I thought (I believed one can only swap numbers, but not digits within the same number) and the computations have to be done again.

I am not sure I follow you. Which estimation, and where leading zeros should be allowed? Thanks!

I am not sure how to continue this, but if we had a functor $F: \textrm{Grp} \to A$, where $A$ is an abelian category, we could infer that $[F(A_4)] = [F(D_6)]$ in the Grothendieck ring of $A$ (which are known in some cases) and see if there are obstructions. I haven't found a satisfying option though. My attempts: homotopy groups of classifying spaces and character rings (which could work, but I am not smart enough to work this out).

Right! I thought that being constrained in a particular arithmetic class does not change the density of primes. But, indeed, since we are excluding multiples of $3$, we have to take into account we discarded a good amount of bad numbers.