My colleague K.-J. Wirths came up with another Rule of Sarrus for ${4\times 4}$ matrices. His suggestion is somehow closeto the original (at least graphically) and is easier to memorize. One has to use the “original” Rule of Sarrus for the ${4\times 4}$ case but now three times. For the first case use the original matrix and for the next two case one has to permute two columns. Graphically this gives the following the pictures:   In principle this generalizes to larger ${n\times n}$ matrices. But beware: ${n!}$ is large! For the ${5\times 5}$ case one has a sum of 120 products but each “standard Sarrus” only gives 10 of them. Hence, one has to figure out 12 different permutations. In the ${n\times n}$ case one even needs to memorize ${\frac{n!}{2n}}$ permutation, let alone all the computations…

I am sure that somebody with stronger background in algebra and more knowledge about permutation groups could easily figure out what is going on here, and to visualize the determinants better.

Update: Indeed! Somebody with more background in algebra already explored how to generalize the Sarrus rule to larger matrices. Again it was my colleague K.-J. Wirths who found the reference and here it is:

• Обобщенное правило Саррюса, by С. Аршон, Матем. сб., 42:1 (1935), 121–128

and it is from 1935 already! If you don’t speak Russian, in German it is

• “Verallgemeinerte Sarrussche Regel”, S. Arschon, Mat. Sb., 42:1 (1935), 121–128

and if you don’t speak German either, you can visit the page in mathnet.ru or to the page in the Zentralblatt (but it seems that there is no English version of the paper or the abstract available…) Anyway, you need $(n-1)!/2$ permutations of the columns and apply the plain rule of Sarrus to all these (and end up, of course, with $2n(n-1)!/2=n!$ summands, each of which has $n$ factors – way more than using LU of QR factorization.)