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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}$-rank. 1 Yes for$\mathbf{Q}$, no for$\mathbf{Q}^*$. For$\mathbf{Q}$, write it as an increasing sequence$L_n$with$L_1=\mathbf{Z}$and$L_n$of finite index over$L_0$. Pick a function$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.