In Exercise 7.3.2 in Weibel's book An Introduction to Homological algebra the following description of the free Lie algebra over some $k$-module $M$ is given (where $k$ is any commutative ring):
First, consider the tensor algebra $T(M)$. Take the underlying Lie algebra $\mathrm{Lie}(T(M))$, i.e. $[x,y] = xy-yx$. Then, consider the Lie subalgebra $\mathfrak{f}(M) \subseteq \mathrm{Lie}(T(M))$ generated by $M$. The claim is that $\mathfrak{f}(M)$ is the free Lie algebra on $M$, i.e. that every $k$-linear map $f : M \to \mathfrak{g}|_k$ into the underlying $k$-module of some Lie algebra extends uniquely to a Lie algebra map $\overline{f} : \mathfrak{f}(M) \to \mathfrak{g}$.
Well, uniqueness is clear, since $M$ generates $\mathfrak{f}(M)$. Specifically, we have to map $[x_1,[x_2,[\dotsc[x_{n-1},x_n]\dotsc]]]$ to $[f(x_1),[f(x_2),[\dotsc[f(x_{n-1}),f(x_n)]\dotsc]]]$ and $\mathfrak{f}(M)|_k$ is generated by these elements. But I am not able to show existence, i.e. that this is well-defined.
By naturality, we may assume $M=\mathfrak{g}|_k$ and $f=\mathrm{id}$. The simplest special case for well-definedness is the following: If $[x,y]=0$ in $\mathfrak{f}(M)$, i.e. $x \otimes y = y \otimes x$ in $M^{\otimes 2}$, why do we have $[x,y]=0$ in $\mathfrak{g}$? This holds when $k$ is a field, because then $x \otimes y = y \otimes x$ implies $x \in \langle y \rangle$ and $[x,y]=0$ follows from $[y,y]=0$. If $k$ is arbitrary, I can only see $2 \cdot [x,y]=0$ (in fact, $[x,y]=[y,x]$ because the Lie bracket is bilinear, and $[y,x]=-[x,y]$ holds in general). So if $2 \in k^{\times}$, it would follow again that $[x,y]=0$, but otherwise it is not clear.
Shouldn't it be possible to construct a counterexample? It doesn't appear in the Errata, though. Is the description true, at least, when $k$ is a field? This should be connected to the PBW-theorem.