The nicest definition of $L_\infty$-algebras ---which I will call a "Chevalley-Eilenberg" style definition after the obvious analogy with the Chevalley-Eilenberg differential of Lie algebras--- is the following:
Let $z^a$ be formal variables of integer degrees $\deg z^a\in\mathbb Z$. They are a basis of some graded vector space $V$ over the reals or complexes. Form the free graded-commutative algebra $S(V)$ as the quotient $$ S(V):=T[z^a]\mod [z^a \otimes z^b - (-1)^{(\deg z^a)(\deg z^b)} z^b\otimes z^a] $$ of the tensor algebra in the variables $z^a$ by the ideal in brackets; in other words odd $z^a$s anticommute with each other. Then the following differential (where $\frac{\partial}{\partial z^b}$ is a left derivative) $$ Q=\sum_{n=1}^{\infty}\frac{1}{n!} C^b_{a_1a_2\dots a_n} z^{a_1}z^{a_2}\cdots z^{a_n} \frac{\partial}{\partial z^b}\,,\qquad C^b_{a_1a_2\dots a_n}\in\mathbb R\, \forall n $$ defines an $L_\infty$-algebra structure on the degree-shifted dual $W=(V[-1])^\star$ iff $Q^2=0$.
Clearly if $\deg z^a=1$ we obtain the definition of a Lie algebra in terms of some set of structure constants $C^b_{a_1a_2}$, and $Q^2=0$ implies the Jacobi identities. In fact $Q$ is the Chevalley-Eilenberg differential in that case. There are other definitions of $L_\infty$-algebras, involving either coalgebras or an infinite number of higher products, neither of which are particularly nice.
In this language, a minimal $L_\infty$-algebra is one with $C^b_{a}=0$ i.e. no linear term in $Q$ (this term defines the dual to the $L_\infty$-algebra unary bracket). The minimal model theorem claims that every $L_\infty$-algebra is quasiisomorphic to a minimal $L_\infty$-algebra (its minimal model); for a construction see e.g. the review in the recent paper 1809.09899.
What I would like to know is if the minimal model has been explicitly constructed in terms of a "Chevalley-Eilenberg" style definition anywhere.
