Suppose $FT$ is the class of all isomorphism classes of finite T-groups (a T-group is a group, where all subnormal subgroups are normal). Define, $\Sigma$ as the set of all functions $f$ from $FT$ to $\mathbb{Z}$, such that $f(E) = 1$, where $E$ is the trivial group. Now, if $f, g \in \Sigma$, define $$(f \ast g) (G)= \Sigma_{H \triangleleft G} f(H)g(\frac{G}{H}).$$
It is not hard to see, that $(\Sigma, \ast)$ is a group:
$f \ast (g \ast h) = \Sigma_{K \triangleleft H \triangleleft G} f(K)g(\frac{H}{K})h(\frac{G}{H}) = (f \ast g) \ast h$
The function $e$, such that $e(E) = 1$ and $e(G) = 0$ for any non-trivial $G$, is the identity element.
The inverse to $f$ is the function $f^{-1}$ satisfying the recurrent relation $\Sigma_{H \triangleleft G} f(H)f^{-1}(\frac{G}{H}) = 0$ and $f^{-1}(E) = 1$.
Moreover, it is quite easy to prove that this group is torsion-free:
Suppose $f \in \Sigma$, $G$ is the nontrivial group of minimal order, such that $f(G) \neq 0$. Then $\forall H \triangleleft G$ if $H \neq G$ and $H \neq E$, then $f(H) = 0$ (as $|H| \leq |G|$). Then $f^n(G) = f(G) + f^{n-1}(G) = nf(G) \neq 0$. That means $\forall n \in \mathbb(N) f^n \neq e$
Personally, I think that this group is also very likely to generate the variety of all groups. However, I do not know how to prove that.
First, I wanted to show, that every marginal subgroup corresponding to a proper variety is trivial in $\Sigma$. However, I failed even to prove that $Z(\Sigma) \cong E$.
Then I thought, that maybe I have to find a subgroup of $\Sigma$ isomorphic to $F_2$ (a free group to two generators). However, I failed in that direction too.
