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In this answer we will focus on the cubic term, which seems to be OP's main question.

1. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $\langle\cdot,\cdot\rangle$ is invariant/associative.

2. Consider fields ${\bf e}$ that are both Lie-algebra-valued, form-valued & supernumber-valued. Note that in OP's references the $n$-forms are (implicitly) interpreted as carrying Grassmann-degree $n$ (modulo 2). The total Grassmann-parity of the fields ${\bf e}$ is assumed to be odd, so that such fields anti-commute. The trilinear form $t$ therefore becomes totally symmetric wrt. such fields.

3. In BV-CS theory (before gauge-fixing), we consider a minimal field $$ {\bf e} ~=~ c ~+~\underbrace{A_{\mu}\mathrm{d}x^{\mu}}_{=~{\bf A}}~+~\underbrace{A^{\ast\mu}(\star \mathrm{d}x)_{\mu}}_{=~{\bf A}^{\ast}} ~+~\underbrace{c^{\ast}\Omega}_{=~{\bf c}^{\ast}} \tag{B}$$ of above type, where $$(\star \mathrm{d}x)_{\mu}~:=~\frac{1}{2}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda}\tag{C}$$ and where $$\Omega~:=~\frac{1}{6}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda} ~=~\frac{1}{3}\mathrm{d}x^{\mu}\wedge(\star \mathrm{d}x)_{\mu}.\tag{D}$$ (The wedges will be not be written explicitly from now on.)

4. The cubic action term is a multinomial expression $$\begin{align} \left. \frac{1}{6} t({\bf e},{\bf e},{\bf e})\right|_{\text{top-form}}~=~& \frac{1}{6}t({\bf A},{\bf A},{\bf A})+ t({\bf A}^{\ast},{\bf A},c) +\frac{1}{2}t({\bf c}^{\ast},c,c)\cr ~=~&\left( t(A_1,A_2,A_3)+ 3t(A^{\ast\mu},A_{\mu},c) +\frac{1}{2}t(c^{\ast},c,c)\right) \Omega.\end{align}\tag{E}$$ Note that in the first line of eq. (E) the (reciprocal) coefficient of each term is precisely its symmetry factor. Eq. (E) agrees with OP's eq. (1).

In this answer we will focus on the cubic term, which seems to be OP's main question.

1. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $\langle\cdot,\cdot\rangle$ is invariant/associative.

2. Consider fields ${\bf e}$ that are both Lie-algebra-valued, form-valued & supernumber-valued. Note that in OP's references the $n$-forms are (implicitly) interpreted as carrying Grassmann-degree $n$ (modulo 2). The total Grassmann-parity of the fields ${\bf e}$ is assumed to be odd, so that such fields anti-commute (in the appropriate graded symmetric tensor algebra). The trilinear form $t$ therefore becomes totally symmetric wrt. such fields.

3. In BV-CS theory (before gauge-fixing), we consider a minimal field $$ {\bf e} ~=~ c ~+~\underbrace{A_{\mu}\mathrm{d}x^{\mu}}_{=~{\bf A}}~+~\underbrace{A^{\ast\mu}(\star \mathrm{d}x)_{\mu}}_{=~{\bf A}^{\ast}} ~+~\underbrace{c^{\ast}\Omega}_{=~{\bf c}^{\ast}} \tag{B}$$ of above type, where $$(\star \mathrm{d}x)_{\mu}~:=~\frac{1}{2}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda}\tag{C}$$ and where $$\Omega~:=~\frac{1}{6}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda} ~=~\frac{1}{3}\mathrm{d}x^{\mu}\wedge(\star \mathrm{d}x)_{\mu}.\tag{D}$$ (The wedges will be not be written explicitly from now on.)

4. The cubic action term is a multinomial expression $$\begin{align} \left. \frac{1}{6} t({\bf e},{\bf e},{\bf e})\right|_{\text{top-form}}~=~& \frac{1}{6}t({\bf A},{\bf A},{\bf A})+ t({\bf A}^{\ast},{\bf A},c) +\frac{1}{2}t({\bf c}^{\ast},c,c)\cr ~=~&\left( t(A_1,A_2,A_3)+ t(A^{\ast\mu},A_{\mu},c) +\frac{1}{2}t(c^{\ast},c,c)\right) \Omega.\end{align}\tag{E}$$ Note that the (reciprocal) coefficient of each term of eq. (E) is precisely its symmetry factor. Eq. (E) agrees with OP's eq. (1).

In this answer we will focus on the cubic term, which seems to be OP's main question.

1. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $\langle\cdot,\cdot\rangle$ is invariant/associative.

2. Consider fields ${\bf e}$ that are both Lie-algebra-valued, form-valued & supernumber-valued. The form-degree and the Grassmann-parity are together assumed to be odd modulo 2 so that such fields anti-commute. The trilinear form $t$ therefore becomes totally symmetric wrt. such fields.

3. In BV-CS theory (before gauge-fixing), we consider a minimal field $$ {\bf e} ~=~ c ~+~\underbrace{A_{\mu}\mathrm{d}x^{\mu}}_{=~{\bf A}}~+~\underbrace{A^{\ast\mu}(\star \mathrm{d}x)_{\mu}}_{=~{\bf A}^{\ast}} ~+~\underbrace{c^{\ast}\Omega}_{=~{\bf c}^{\ast}} \tag{B}$$ of above type, where $$(\star \mathrm{d}x)_{\mu}~:=~\frac{1}{2}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda}\tag{C}$$ and where $$\Omega~:=~\frac{1}{6}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda} ~=~\frac{1}{3}\mathrm{d}x^{\mu}\wedge(\star \mathrm{d}x)_{\mu}.\tag{D}$$ (The wedges will be not be written explicitly from now on.)

4. The cubic action term is a multinomial expression $$\begin{align} \left. \frac{1}{6} t({\bf e},{\bf e},{\bf e})\right|_{\text{top-form}}~=~& \frac{1}{6}t({\bf A},{\bf A},{\bf A})+ t({\bf A}^{\ast},{\bf A},c) +\frac{1}{2}t({\bf c}^{\ast},c,c)\cr ~=~&\left( t(A_1,A_2,A_3)+ 3t(A^{\ast\mu},A_{\mu},c) +\frac{1}{2}t(c^{\ast},c,c)\right) \Omega.\end{align}\tag{E}$$ Note that in the first line of eq. (E) the (reciprocal) coefficient of each term is precisely its symmetry factor. Eq. (E) agrees with OP's eq. (1).

In this answer we will focus on the cubic term, which seems to be OP's main question.

1. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $\langle\cdot,\cdot\rangle$ is invariant/associative.

2. Consider fields ${\bf e}$ that are both Lie-algebra-valued, form-valued & supernumber-valued. Note that in OP's references the $n$-forms are (implicitly) interpreted as carrying Grassmann-degree $n$ (modulo 2). The total Grassmann-parity of the fields ${\bf e}$ is assumed to be odd, so that such fields anti-commute. The trilinear form $t$ therefore becomes totally symmetric wrt. such fields.

3. In BV-CS theory (before gauge-fixing), we consider a minimal field $$ {\bf e} ~=~ c ~+~\underbrace{A_{\mu}\mathrm{d}x^{\mu}}_{=~{\bf A}}~+~\underbrace{A^{\ast\mu}(\star \mathrm{d}x)_{\mu}}_{=~{\bf A}^{\ast}} ~+~\underbrace{c^{\ast}\Omega}_{=~{\bf c}^{\ast}} \tag{B}$$ of above type, where $$(\star \mathrm{d}x)_{\mu}~:=~\frac{1}{2}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda}\tag{C}$$ and where $$\Omega~:=~\frac{1}{6}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda} ~=~\frac{1}{3}\mathrm{d}x^{\mu}\wedge(\star \mathrm{d}x)_{\mu}.\tag{D}$$ (The wedges will be not be written explicitly from now on.)

4. The cubic action term is a multinomial expression $$\begin{align} \left. \frac{1}{6} t({\bf e},{\bf e},{\bf e})\right|_{\text{top-form}}~=~& \frac{1}{6}t({\bf A},{\bf A},{\bf A})+ t({\bf A}^{\ast},{\bf A},c) +\frac{1}{2}t({\bf c}^{\ast},c,c)\cr ~=~&\left( t(A_1,A_2,A_3)+ 3t(A^{\ast\mu},A_{\mu},c) +\frac{1}{2}t(c^{\ast},c,c)\right) \Omega.\end{align}\tag{E}$$ Note that in the first line of eq. (E) the (reciprocal) coefficient of each term is precisely its symmetry factor. Eq. (E) agrees with OP's eq. (1).

Hint: There is 6 (3) combinations of the multi-nomial expression $e^3=(c+A+A^*+c^*)^3$ that yields the $A^*cA$ ($c^*cc$) action term, respectively. This explains all coefficients up to signs.

In this answer we will focus on the cubic term, which seems to be OP's main question.

1. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $\langle\cdot,\cdot\rangle$ is invariant/associative.

2. Consider fields ${\bf e}$ that are both Lie-algebra-valued, form-valued & supernumber-valued. The form-degree and the Grassmann-parity are together assumed to be odd modulo 2 so that such fields anti-commute. The trilinear form $t$ therefore becomes totally symmetric wrt. such fields.

3. In BV-CS theory (before gauge-fixing), we consider a minimal field $$ {\bf e} ~=~ c ~+~\underbrace{A_{\mu}\mathrm{d}x^{\mu}}_{=~{\bf A}}~+~\underbrace{A^{\ast\mu}(\star \mathrm{d}x)_{\mu}}_{=~{\bf A}^{\ast}} ~+~\underbrace{c^{\ast}\Omega}_{=~{\bf c}^{\ast}} \tag{B}$$ of above type, where $$(\star \mathrm{d}x)_{\mu}~:=~\frac{1}{2}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda}\tag{C}$$ and where $$\Omega~:=~\frac{1}{6}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda} ~=~\frac{1}{3}\mathrm{d}x^{\mu}\wedge(\star \mathrm{d}x)_{\mu}.\tag{D}$$ (The wedges will be not be written explicitly from now on.)

4. The cubic action term is a multinomial expression $$\begin{align} \left. \frac{1}{6} t({\bf e},{\bf e},{\bf e})\right|_{\text{top-form}}~=~& \frac{1}{6}t({\bf A},{\bf A},{\bf A})+ t({\bf A}^{\ast},{\bf A},c) +\frac{1}{2}t({\bf c}^{\ast},c,c)\cr ~=~&\left( t(A_1,A_2,A_3)+ 3t(A^{\ast\mu},A_{\mu},c) +\frac{1}{2}t(c^{\ast},c,c)\right) \Omega.\end{align}\tag{E}$$ Note that in the first line of eq. (E) the (reciprocal) coefficient of each term is precisely its symmetry factor. Eq. (E) agrees with OP's eq. (1).