One dimension ({0,1} n*n matrix with exactly one 1 per row or column): The number of permutations satisfies Stirling's formula: it's [(1+o(1)(n/e)]^n

Two dimensions ({0,1} n*n*n matrix with exactly one 1 per row, column or shaft): The number of Latin squares, as seen from Van Lint and Wilson's book, is [(1+o(1))(n/e^2)]^{n^2}.

Conjecture: in d dimensions, it's [(1+o(1))(n/e^d)]^{n^d}

Linial and Zur Luria proved the "less than or equal to" side. The "greater than or equal to" is still open.