Here is a proof that the statement is, as you expected, false (for $n\ge 2$), in case the field $K$ is $\mathbf{R}$ or $\mathbf{C}$. Unfortunately it does not give anything explicit.

Let $B$ be its closed unit ball. Then $\hat{L}_n$ is a Polish topological group.

If by contradiction, $\hat{L}_n$ is group-wise generated by $\bigcup_{i=1}^n Kx_i$, then it is also generated by the compact symmetric subset $S=\bigcup_{i=1}^n Bx_i$. So $\hat{L}_n=\bigcup_m S^m$. By Baire's theorem, there exists $m$ such that the compact subset $S^m$ has nonempty interior. So the topological group $\hat{L}_n$ is locally compact, contradiction. (In general, a Polish group generated by a compact subset is locally compact: this is well-known.)

Added: this is a contradiction simply because $\hat{L}_n$ is not locally compact (for $n\ge 2$) and this is something which can be seen directly. Indeed, if $V$ is a neighborhood of 0, then it contains the kernel $U_m=(\hat{L}_n)_{\ge m}$ for some $m$ (which is the $m$-term in the lower central series). Then choosing nonzero $x$ in $U_m$ and considering its scalar multiples, we see that $U_m$ does not have compact closure ($U_m$ is closed, so I'm just saying $U_m$ is not compact). Hence $\hat{L}_n$ is not locally compact.

The argument also works if $K$ can be made a Polish ring for which it is a countable union of compact subsets (say $B_m$ with $0\in B_m\subset B_{m+1}$). In this case the same argument works using the exhaustion by the $(\bigcup_{i=1}^nB_mx_i)^m$.

This applies to $K$ countable (well, obvious since the $\bigcup Kx_i$ is countable), and also to $K$ $p$-adic field.