Let me answer the question in the title rather than in the body. First of all for the free profinite group on one generator $\widehat Z$ we have a direct product of p-adic integers over all primes and so we know how to write down elements. Now if $w$ is any element of a profinite group and $\lambda\in \widehat Z$ then define $w^\lambda$ to be the image of $\lambda$ under the map sending the generator of $\widehat Z$ to $w$. So we can build up some elements by starting with words and closing under products and infinite powers.
Here is an example. Let $p$ be a prime. The sequence $p^{n!}$ converges in $\widehat Z$ to a generator, denoted $p^{\omega}$, of the $p'$-component (i.e., the product of all q-adic groups with $q\neq p$). Hence a profinite group is pro-$p$ if and only if it satisfies the profinite identity $x^{p^{\omega}}=1$ because $x^{p^{\omega}}$ generates the p'-prime component of the pro-cyclic subgroup generated by $x$.
Jorge Almeida, principally in the context of free profinite monoids, but also for free profinite groups, came up with the following method to generate further elements. Let $M=End(\widehat F_r)$ be the endomorphism monoid. It is a profinite monoid. Given any endomorphism $\phi$ one has that $\phi^{\omega}=\lim_{n\to \infty} \phi^{n!}$ is an idempotent endomorphism. One can create interesting and useful elements which are recursive (in several senses, eg,there is an re sequence converging to it and one can compute its image in any finite quotient group) as follows.
Take an endomorphism $\phi$ of $F_r$. The take a generator $x$ of $F_r$ and consider the element $\phi^{\omega}(x)$. This element is easily computed in the above senses and can be interesting. For example $\phi(x)=x^p$ is an endomorphism of the free group on one generator. The map $\phi^{\omega}$ takes $x$ to $x^{p^{\omega}}$.
Define $\psi\colon F_2\to F_2$ by $\psi(x)=[x,y]$ and $\psi(y)=y$ where $x,y$ are free generators. Then $\psi^{\omega}(x)$ is like an infinite iterated commutator $[x,y,y,...]$. Almeida denotes this element $[x,_{\omega}y]$ and observes a profinite group is pro-nilpotent iff it satisfies the profinite identity $[x,_{\omega} y]=1$.
There is also a 2-variable profinite identity defining pro-solvable groups, but this is more complicated and if memory serves relies on Thompson's classification of minimal non-sovable groups and maybe even the classification of finite simple groups.
Here are some Almeida surveys.
- http://cmup.fc.up.pt/cmup/jalmeida/preprints/finite-semigroups-and-dynamics-pdf.pdf
- http://cmup.fc.up.pt/cmup/jalmeida/preprints/Engel-CMUP.pdf
Best Answer
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.