monomials in the universal enveloping of a Lie algebra in terms of the symmetric basis Let $L$ be a finite-dimensional Lie algebra over a field $k$ of characteristic zero and $e_1,\ldots, e_n$ some basis of $L$. The formula $[e_i,e_j] = \sum_k C_{ij}^k e_k$ determines the structure coefficients $C_{ij}^k$. Given any ordered $k$-tuple $I = (i_1,\ldots,i_k)\in \lbrace 1,\ldots,n \rbrace^k$, define $e_I = e_{i_1}\cdots e_{i_k}\subset U(L)$ and 
$$
e^S_I = \frac{1}{n!}\sum_{\sigma\in\Sigma(k)} e_{\sigma(i_1)}\cdots e_{\sigma(i_k)}\in U(L).
$$
As it is well known, from various forms of a PBW theorem, $e_I$, for all $I$ with $i_1\lt i_2\lt \ldots \lt i_k$, $k \geq 0$, form a basis and also $e^S_I$ (for the same set of $I$-s) form a basis. I need explicit formulas for $e_I$ in the linear basis of $e^S_J$-s where the coefficients are expressed in terms of the structure constants $C_{ij}^k$ (and combinatorial factors). In fact, for my present purposes, it would be enough to know explicitly the deepest, lowest order, linear term (but it is of course the hardest summand in the expansion).  For example, for the easiest nontrivial case $k = 2$, 
$$e_{(i,j)} = e_i e_j = \frac{1}{2}(e_i e_j + e_j e_i) + \frac{1}{2} \sum_k C_{ij}^k e_k = e_{(i,j)}^S + \frac{1}{2} \sum_k C_{ij}^k e_k,$$
hence the linear term is $\frac{1}{2} \sum_k C_{ij}^k e_k$.
 A: An explicit formula is given in this paper by L. Solomon. I copy the abstract here:

Let g be a Lie algebra over a field of characteristic zero. Let T be the tensor algebra of g, let S be the subspace of symmetric tensors and let J be the two-sided ideal of T generated by tensors x⊗y−y⊗x−[x, y]. One formulation of the P-B-W theorem states that T=S⊕J, direct sum. In this paper we give an explicit formula for the projection of T on S defined by this direct decomposition.

A: This will not be an honest answer but just a long remark: it reminds me a bit on the relation between Weyl quantization and standard ordered quantization. Here one has the polynomials in $q$ and $p$ with their canonical Poisson bracket $\lbrace q, p\rbrace = 1$ which should be quantized into operators as usual. For higher polynomials one has to choose an ordering, e.g. Weyl or standard or many more... Pulling back the operator product gives then a star product, depending on the choice of the ordering. All of them are isomorphic by explicit isomorphism. In case of Weyl/standard the isomorphism is given by the exponential of the indefinite Laplacian $\frac{\partial^2}{\partial q \partial p}$. This is very explicit and allows for many nice formulas and computations.
Now in your situation it seems to me that you would like to have some similar isomorphism between the two quantizations of the Polynomials on the dual $\mathfrak{g}^*$ which you obtain by total symmetrization (= Weyl) and a standard ordering with respect to the choice of a basis of $\mathfrak{g}$. The point is that the standard ordering you are considering is much less canonical as in the case of canonical quantization: you really have to specify a basis.
I don't think that this has been worked out in detail, but maybe the old work of Simone Gutt from 1983 in Lett. Math. Phys. might be inspiring.
EDIT: Maybe I have misunderstood the question in the first place. So if your intention is to find an explicit formula for the Weyl ordered case alone, then one indeed has an "explicit" formula: I'm not sure where this showed up first, but the formula goes as follows: given $\xi, \eta \in \mathfrak{g}$ one considers the (formal) BCH series $BCH(\hbar \xi, \hbar\eta)$ with $\hbar$ being a formal parameter. Then the Weyl product of the formal exponentials $\exp(\hbar\xi)$ and $\exp(\hbar\eta)$ is given by
$$\exp(\hbar\xi) \star \exp(\hbar\eta)
=\exp (\hbar^{-1} 
 BCH(\hbar \xi, \hbar\eta))
$$
Well, from this one get's the formula for monomials by differentiating and polarization. But you see, you won't get what you want: this is not at all explicit for two reasons. First, the differentation and polarization has to be done, very ugly. But second, and this is the more severe point, you have to know the BCH series.
Now on the other hand, this formula shows that you probabaly can not expect to get a simpler formula without using BCH. I fear, one can not go beyond this... :(
