As seen in Section 4 of Chapter 5 of Costello, K. "Renormalization and Effective Field Theory", or in section 5.2 [$L_\infty$-Algebras of Classical Field Theories and the Batalin-Vilkovisky Formalism][1], the BV form of the Chern-Simons action is
$$S=\frac{1}{2}\langle A,dA\rangle+\frac{1}{6}\langle A,[A\wedge A]\rangle+\langle A^*,D_Ac\rangle+\frac{1}{2}\langle c^*,[c,c]\rangle,$$
with $c\in\Omega^0(M)\otimes\mathfrak{g}[1]$, $A\in\Omega^1(M)\otimes\mathfrak{g}$, $A^*\in\Omega^2(M)\otimes\mathfrak{g}[-1]$, and $c^*\in\Omega^3(M)\otimes\mathfrak{g}[-2]$. In here $\mathfrak{g}$ is a Lie algebra equipped with an invariant non-degenerate pairing $\langle\cdot,\cdot\rangle$. However, in the first reference it is also claimed that this action can be put into the form
$$S=\frac{1}{2}\langle e,de\rangle+\frac{1}{6}\langle e,[e\wedge e]\rangle$$
for some field $e$. I don't see how this is possible.

Let me explain my reasoning. Let us first assume $e=c+A+A^*+c^*$. Note that $\langle\alpha,\beta\rangle=0$ if $\alpha\in\Omega^p(M)\otimes\mathfrak g$ and $\beta\in\Omega^q(M)\otimes\mathfrak g$ with $p+q\neq 3$. We can use this to expand $\langle e,d{e}\rangle$. For example, the only term that can be coupled with the $A$ coming from the left $e$ is the $d{A}$ coming from $d{e}$. We conclude that
\begin{equation}
    \frac{1}{2}\langle e,d{e}\rangle=\frac{1}{2}\langle c,d{A^*}\rangle+\frac{1}{2}\langle A,d{A}\rangle+\frac{1}{2}\langle A^*,d{c}\rangle.
\end{equation}
Now, remembering that $A^*$ and $c$ are fermionic, we have
\begin{equation}
    \begin{aligned}
    \langle c,d{A^*}\rangle&=\int c^ad{A^{*b}}\langle T_a,T_b\rangle_{\mathfrak g}=-\int d{A^{*b}}c^a\langle T_a,T_b\rangle_{\mathfrak g}\\
    &=-\int d{(A^{*b}c^a)}\langle T_a,T_b\rangle_{\mathfrak g}+\int A^{*b}d{c^a}\langle T_a,T_b\rangle_{\mathfrak g}.
    \end{aligned}
\end{equation}
Thus, up to total derivatives we have
\begin{equation}
    \frac{1}{2}\langle e,d{e}\rangle=\frac{1}{2}\langle A,d{A}\rangle+\langle A^*,d{c}\rangle.
\end{equation}

To expand the term $\langle e,[e\wedge e]\rangle$, note that $[e\wedge e]$ can only have even forms. Indeed, an odd form in the expansion of $[e\wedge e]$ must come from the coupling $[\alpha\wedge\beta]$ of an odd form $\alpha$ and an even form $\beta$ in $e$. Since they are different, the term $[\beta\wedge\alpha]$ also appears in the expansion of $e$. Now, all even forms in $e$ are fermionic while all odd forms in $e$ are bosonic. We conclude that $\alpha$ is bosonic while $\beta$ is fermionic. Therefore
\begin{equation}
    [\alpha\wedge\beta]=\alpha^a\wedge\beta^b[T_a,T_b]=\beta^b\wedge\alpha^a[T_a,T_b]=-\beta^b\wedge\alpha^a[T_b,T_a]=-[\beta\wedge\alpha].
\end{equation}
Therefore the terms $[\alpha\wedge\beta]$ and $[\beta\wedge\alpha]$ cancel. By the same token, the rest of the surviving terms in the expansion of $[e\wedge e]$ are symmetric $[\alpha\wedge\beta]=[\beta\wedge\alpha]$. Given that we are in three dimensions, they have to either be 0-forms or 2-forms. We conclude that
\begin{equation}
    [e\wedge e]=[c\wedge c]+2[c\wedge A^*]+[A\wedge A].
\end{equation}
Of course, for $0$-forms we have $[c\wedge c]=[c,c]$. The second term is then  
\begin{equation}
    \frac{1}{6}\langle e,[e\wedge e]\rangle=\frac{1}{6}\langle A,[A\wedge A]\rangle+\frac{1}{3}\langle A,[c\wedge A^*]\rangle+\frac{1}{6}\langle c^*,[c\wedge c]\rangle.
\end{equation}

We see that we have failed to recover our original action because of some factors. One could try to resolve this by combining the fields in $e$ with different numerical factors. However, since the action of $A$ already has the correct factors, we cannot rescale $A$. Indeed, any rescaling of $A$ would produce a mismatch in the scales of the quadratic and cubic terms in $A$. On the other hand, the term $\langle A^*,dc\rangle$ has also the correct factor, so that we must scale $c$ and $A^*$ inversely. This means that we will never get the correct factor for the cubic term in $c$, $A$, and $A^*$.


  [1]: https://arxiv.org/abs/1809.09899