First, let me expand on the reply of Dietrich Burde: I got hold of the paper of Hazewinkel, and can now be more precise about what is and what is not there (last time I saw it was some years ago).

Hazewinkel's most general result (for not necessarily trivial coefficients) is over a field. A result that he proves over a ring is as follows:

Let $R$ be a hereditary ring, $\mathfrak{g}$ an Lie algebra over $R$ which is a free $R$-module of rank $n$, and $M$ a finitely generated $\mathfrak{g}$-module which is projective as an $R$-module. Then there exists a canonical splitting exact sequence
$$
0\to\mathrm{Ext}^1(H^{s+1}(\mathfrak{g},M),R)\to H^{n-s}(\mathfrak{g},(M^{\mathrm{tw}})^*)\to H^s(\mathfrak{g},M)^*\to 0
$$
where the $\mathfrak{g}$-module $M^{\mathrm{tw}}$ has the same underlying $R$-module, but the action on it is given by $\rho^{\mathrm{tw}}(x)=\rho(x)-\mathop{\mathrm{tr}}(\mathop{\mathrm{ad}}(x))\cdot1$.

In the case of $M=R$, $(M^{\mathrm{tw}})^*$ is $R$ with the $\mathfrak{g}$-action given by multiplying by trace of the adjoint action. The example of YCor in the comment to the original question, in particular, is addressed by this, of course (there, incidentally, $R^{\mathrm{tw}}\simeq R$ as $\mathfrak{g}$-modules): we have $H^1(\mathfrak{sl}_2,\mathbb{Z})=0$ and $H^2(\mathfrak{sl}_2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$, also we have $H_1(\mathfrak{sl}_2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$, and applying Hazewinkel's result for $s=1$ (so $s+1=2$), we see no contradiction with $\mathrm{Ext}^1(H^{s+1}(\mathfrak{sl}_2,\mathbb{Z}),\mathbb{Z})=\mathrm{Ext}^1((\mathbb{Z}/2\mathbb{Z})^2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$.

However, for the kind of information you seek (Poincaré-type duality between homology and cohomology), the answer is simpler, since the main reason for the universal coefficients-kind phenomena in the formulas of Hazewinkel is that we need to dualise something we do not want to dualise :)

Let me actually first give an example similar to the one Mariano gave, it is a nice simple exercise anyway, and it shows once again that twisting coefficients is unavoidable. Let $\mathfrak{g}$ be a Lie algebra over $\mathbb{Z}$ with two basis elements $x,y$ and the bracket $[x,y]=2y$. Then
$$
H_i(\mathfrak{g},\mathbb{Z})=\begin{cases}\mathbb{Z}, i=0,\\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, i=1,\\ 0, i>1,\end{cases}
$$
$$
H_i(\mathfrak{g},\mathbb{Z}^{\mathrm{tw}})=\begin{cases}\mathbb{Z}/2\mathbb{Z}, i=0,\\ \mathbb{Z}, i=1,\\ \mathbb{Z}, i=2,\\ 0, i>2,\end{cases}
$$
$$
H^i(\mathfrak{g},\mathbb{Z})=\begin{cases}\mathbb{Z}, i=0,\\ \mathbb{Z}, i=1,\\ \mathbb{Z}/2\mathbb{Z}, i=2,\\ 0, i>2,\end{cases}
$$
$$
H^i(\mathfrak{g},\mathbb{Z}^{\mathrm{tw}})=\begin{cases}0, i=0,\\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, i=1,\\ \mathbb{Z}, i=2,\\ 0, i>2,\end{cases}
$$

This example makes one think that we must have $$H_i(\mathfrak{g},R)\simeq H^{n-i}(\mathfrak{g},R^{\mathrm{tw}})\quad \text{ and }\quad H_i(\mathfrak{g},R^{\mathrm{tw}})\simeq H^{n-i}(\mathfrak{g},R).$$ Indeed, this is true because the pairs of complexes computing that (co)homology are isomorphic. Basically, take some basis $e_1\ldots,e_n$ of $\mathfrak{g}$, and send each multivector $e_{i_1}\wedge\cdots\wedge e_{i_k} (i_1<\dots<i_k)$ to the skew-symmetric multilinear function $\phi$ for which, whenever $j_1<\dots<j_l$, we have $\phi(e_{j_1},\ldots,e_{j_l})=\delta_{I,\{1,\ldots,n\}\setminus J}$. This gives an isomorphism of the Chevalley-Eilenberg complexes computing the respective (co)homology groups. (The computation that this takes one differential into another (up to a sign) needs some careful bookkeeping but is quite straightforward, and is essentially contained between the lines in Hazewinkel's paper).

arbitrary fields(for $\mathfrak{g}$ unimodular) is Koszul's paper (quoted in Dietrich Burde's answer), §III.7. $\endgroup$ – YCor Apr 14 at 18:28