I believe an approach that works is to define the Chevalley-Eilenberg complex as a kind of `Koszul complex over the ring of functions'. The enveloping algebra $U$ is relatively quadratic over the ring of functions $O$, and so you can define a quadratic dual algebra $U^\perp$ in analog with the absolute case (when there is a field in degree zero). The corresponding Koszul boundary then squares to zero by a simple argument, and a slightly harder argument shows the Koszul complex is a resolution of $O$ as a left $U$-module.
I worked this out in my paper on the Beilison equivalence for Lie algebroids. The section you want is the first couple of pages of section 4.
EDIT: After rereading your question, it sounds like you are more interested in how to recover the standard Lie algebroid structure (the 'bracket' and 'anchor map') from the CE complex. These should be gotten directly from the zero degree differential: $$ O\rightarrow L^* $$ and the first degree differential $$ L^* \rightarrow L^* \wedge L^* $$ The dual of the first map gives a derivation for every element of $L$ by the Leibniz rule, so it defines the anchor map $L\rightarrow T_X$. The dual of the second map gives a map $L\wedge L \rightarrow L$ which defines the Lie bracket on $L$.
The condition that the $d$ satisfies the Leibniz rule for the product of a degree zero element and a degree one element in $\Lambda^\bullet_X L$ is gives the Leibniz condition for the Lie algebroid, and the condition that $d^2$ is zero on degree 1 elements gives the Jacobi identity.
Is this the kind of answer you were looking for?
Oh, I should mention that this argument, the nLab page, and my paper assume $L$ is a vector bundle; some other sources might not. The result you mention (the differential on $\Lambda^\bullet L^*$ being equivalent to a Lie algebroid structure) is not true in that context.