Yes for $\mathbf{Q}$, no for $\mathbf{Q}^*$.
For $\mathbf{Q}$, write it as the union of an increasing sequence $L_n$ with $L_1=\mathbf{Z}$ and $L_n$ of finite index over $L_0$. L_1$. Pick a function $F$ with fast growth and define $l'(r)=|r|+F(\sup{n:r\notin L_n})$.
For $\mathbf{Q}^*$, it contains a subgroup isomorphic to $\mathbf{Z}^d$ for every $d$, so for every length the growth is at least polynomial of degree $d$. So it can't be polynomial.
Added-1: for $\mathbf{Q}$ we can arrange so that $[L_n:L_1]\le n$ for all $n$. Then we can pick $F$ to be the identity and then the growth is at most quadratic.
Added-2: the argument extends, showing that an abelian group admits a length with polynomial growth iff it's countable and has finite $\mathbf{Q}$-rank.
