Not that I am happy with the forthcoming proof, but at least it is human.
As Iosif Pinelis notes, both sides of your equation are multiaffine. Also, for every $i=1,2,3$ they are homogeneous of degree 2 with respect to $R_i:=(a_i,b_i,c_i,d_i)$. Thus, every monomial which may appear after expanding the determinants has 2 variables from each of sets $R_1,R_2,R_3$. How to get out a coefficient of monomial $\prod_{i\in S} x_i$ in the homogeneous multiaffine polynomial $p(x_1,x_2,\ldots,x_n)$, where $S\subset \{1,\ldots,n\}$? Just put $x_i=1$ for $i\in S$ and $x_i=0$ for $i\notin S$, the value of this polynomial is the desired coefficient.
Thus, it suffices to check your identity when two guys in each $R_i$ are equal to 1 and two guys are equal to 0. In particular, this yields $a_i(b_i+c_i+d_i)=a_i$ etc. So, your three determinants in LHS now surprisingly have the same first three rows $R_1,R_2,R_3$ as the determinant in RHS. Using the linearity of the determinant with three fixed rows as the function of the fourth row, we may rewrite now your identity as
$$\begin{vmatrix}
a_1 & b_1 & c_1 & d_1 \\
a_2 & b_2 & c_2 & d_2 \\
a_3 & b_3 & c_3 & d_3 \\
a_4 & b_4 & c_4 & d_4
\end{vmatrix}=0,\, \text{where}\, a_4=a_1a_2+a_2a_3+a_3a_1-3a_1a_2a_3,\,\text{etc}.$$
Note that $a_4=0$ unless exactly two entries of $A:=(a_1,a_2,a_3)$ are equal to 0 (in this case $a_4$ equals 1), analogously for other columns. Therefore, if no column and no row of our $4\times 4$ matrix is 0, each sequence $A,B,C,D$ must contain at least one 1, and at least one of them must contain exactly two 1's. Since totally they contain 6 1's, we conclude that two of them (say, $A$ and $B$) contain two 1's. There exists $i\in \{1,2,3\}$ such that $a_i=b_i=1$. Then the $i$-th and 4-th lines of our matrix coincide.
Best Answer
The answer is no. Indeed, let $T:=\sigma_3=(t_{ij}\colon i,j\in\{1,2,3\})$, $A:=A_3$, and $$B:=\begin{pmatrix} a & -b_2 & 0 \\ -b_1 & a & -c_2 \\ 0 & -c_1 & a \end{pmatrix}.$$ Then the equality in question would imply that $$TA=BT \tag{1} \label{1}$$ for all $a,b_1,b_2,c_1,c_2$. Solving system \eqref{1} of linear equations for the $t_{ij}$'s, we get $$b_1 t_{11}+c_2 t_{13}+b_2 t_{22}=0,\quad c_1 t_{12}+b_2 t_{23}=0$$ for all $b_1,b_2,c_1,c_2$. It follows that $$t_{11}=t_{13}=t_{22}=0,\quad t_{12}=t_{23}=0,$$ so that $t_{11}=t_{13}=t_{12}=0$ and hence $\det T=0$, which precludes the desired identity $TAT^{-1}=B$. $\quad\Box$
In fact, one can see that, if \eqref{1} holds for all $a,b_1,b_2,c_1,c_2$, then necessarily $T=0$.