Python 3Python 3, 238 bytes, 227 bytes, 224 bytes, 216 bytes

from functools import*
from itertools import*
r=range;n=len;s=sum
f=lambda l:s(reduce(lambda p,m:p*m,[l[a][b]for a,b in zip(r(n(l)),j)])*(-1)**s(s(y<j[x]for y in j[x:])for x in r(n(l)))for j in permutations(r(n(l))))

from functools import*
from itertools import*
r=range;n=len;s=sum
f=lambda l:s(reduce(lambda p,m:p*m,[l[a][b]for a,b in zip(r(n(l)),j)])*(-1)**s(s(y<j[x]for y in j[x:])for x in r(n(l)))for j in permutations(r(n(l))))

Try it online!Try it online!

My solution uses the definition of a determinant for calculations. Unfortunately, the complexity of this algorithm is n! and I can not show the passage of the last test, but in theory this is possible.

Python 3, 238 bytes, 227 bytes, 224 bytes, 224216 bytes

from functools import*
from itertools import*
r=range;n=len;s=sum
f=lambda l:sums(reduce(lambda p,m:p*m,[l[a][b]for a,b in zip(ranger(lenn(l)),j)])*(-1)**(sum(**s(sums(y<j[x]for y in j[x:])for x in ranger(lenn(l)))))for j in permutations(ranger(lenn(l))))

Try it online!

My solution uses the definition of a determinant for calculations. Unfortunately, the complexity of this algorithm is n! and I can not show the passage of the last test, but in theory this is possible.

Python 3, 238 bytes, 227 bytes, 238224 bytes

from operator import*
from functools import*
from itertools import*
f=lambda l:sum(reduce(mul,[l[a][b]lambda forp,m:p*m,[l[a][b]for a,b in zip(range(len(l)),j)])*(-1)**(sum((sum(y<j[x]for y in j[x:]) for x in range(len(l))))) for j in permutations(range(len(l))))

Try it online!

My solution uses the definition of a determinant for calculations. Unfortunately, the complexity of this algorithm is n! and I can not show the passage of the last test, but in theory this is possible.