[[There is some overlap with Theo's answer but I have not tried to take that into consideration. [Except now for the addendum.] ]]
Well, now that we have been placed ourselves in characteristic zero we can, in the usual case, use the canonical isomorphism $S^\ast\mathfrak{g}\rightarrow U(\mathfrak{g})$ given by the composite $S^n\mathfrak{g}\subseteq \mathfrak{g}^{\otimes n} \rightarrow U(\mathfrak{g})$, with the last map given by multiplication. This imbues $S^\ast\mathfrak{g}$ with a funny multiplication and if we can express it using only the Lie bracket and tensor concatenation we can extend it to a general symmetric monoidal category context. The twisted multiplication is given by a bunch of maps $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \rightarrow S^k\mathfrak{g}$ (where it is easy to see that the maps are zero unless $k\leq i+j$). It is enough to consider the $n$-product map $\mathfrak{g}^{\otimes n}\rightarrow S^k\mathfrak{g}$ is the general map is the composite of $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \subseteq \mathfrak{g}^{\otimes i+j} \rightarrow S^k\mathfrak{g}$. To get the idea that there are universal formulas we look at small values: We have $u\otimes v=u\odot v+1/2[u,v]$, where $u\odot v$ is the symmetrisation $1/2(u\otimes v+v\otimes u)$. Similarly, considering $u\odot v\odot w$ we may systematically transpose the summands distinct from $u\otimes v\otimes w$ at the expense of putting in a commutator to get (for instance) $$ u\odot v\odot w = u\otimes v\otimes w + \frac{1}{2}u\otimes [w,v]+v\otimes[w,u]+w\otimes[v,u]+\frac{1}{3}([u,v]\otimes v+[u,w]\otimes w). $$ We can then recursively reduce the rest of the terms. (Note that the exact formulas depend exactly how we perform the reduction.) This continues in higher degrees and give specific (making as I said some specific choices for the reductions but the Jacobi identity and the anti-symmetry imply that the result is independent of those choices). we can now transfer these formulas to a general symmetric monoidal category.
It seems not at all clear that you get an associative algebra in this way but I think one should be able to prove that it is a formal consequence of the anti-symmetry and the Jacobi identity by considering the free Lie algebra on varying vector spaces.
Addendum: As Theo points out, the enveloping algebra should preferably have a universal property. Giving a Lie algebra morphism $\mathfrak{g} \rightarrow A$ into an associative algebra object we can get maps $S^i\mathfrak{g} \rightarrow A$ by composing the inclusion $S^n\mathfrak{g}\subseteq \mathfrak{g}^{\otimes n}$, the induced map $\mathfrak{g}^{\otimes n} \rightarrow A^{\otimes n}$ and the product map of $A$. That this gives an algebra map $U(\mathfrak{g}) \rightarrow A$ comes down to some universal identities which again one should be able to verify by reduction to the ordinary case.