The logic behind this does extend to general $n\times n$ determinants, though probably not as nicely as you wish. Note that I am taking the liberty to replace "last" by "first" in "where R cyclically permutes the last three rows of the matrix A". It doesn't matter, because Sarrus' rule is invariant under cyclic shift, and a simple cyclic shift turns the last three rows to the first three rows.
Consider the alternating group $A_{n-1}$ embedded into the symmetric group $S_n$: every element of $A_{n-1}$ is a permutation of the set $\left\lbrace 1,2,...,n-1\right\rbrace$, and thus can be seen as a permutation of the set $\left\lbrace 1,2,...,n\right\rbrace$ which leaves $n$ fixed.
Also consider the dihedral group $D_n$ defined as the subgroup of $S_n$ generated by the cyclic shift $\left(x\mapsto x+1\mod n\right)$ and the reflection $\left(x\mapsto n+1-x\right)$.
Then, every element $\pi\in S_n$ can be uniquely written as $\pi=\sigma\xi$ with $\sigma\in A_{n-1}$ and $\xi\in D_n$. In fact, $\xi$ is uniquely determined by the conditions $\left(\pi\xi^{-1}\right)\left(n\right)=n$ and $\mathrm{sign}\left(\pi\xi^{-1}\right)=1$, and then $\sigma$ results.
Now, write the determinant of an $n\times n$ matrix in the form $\sum_{\pi\in S_n}\mathrm{sign}\pi\cdot\prod ...=\sum_{\sigma\in A_{n-1}}\sum_{\xi\in D_n}\mathrm{sign}\xi\cdot\prod ...$. Each inner sum $\sum_{\xi\in D_n}\mathrm{sign}\xi\cdot\prod ...$ is the naive "Sarrus determinant" of some permutation of the matrix; which permutation it actually is is decided by the $\sigma$.
For $n=3$, we have $A_{n-1}=A_2=1$, so the outer sum $\sum_{\sigma\in A_{n-1}}$ has only one term, and the "Sarrus determinant" is the real determinant.
For $n=4$, we have $A_{n-1}=A_3=C_3$ (the cyclic group with $3$ elements), so the outer sum $\sum_{\sigma\in A_{n-1}}$ has three terms, and it follows that the determinant of a $4\times 4$ matrix can be written as a sum of three "Sarrus determinants". A closer look at the sum shows which ones.
