Estimates of the number of magic squares, cubes, ... (hypercubes)
 Multimagic squares

The E-number can be defined in the same way as for regular magic squares by using the number of multimagic series.
Let's assume that the C-factor for multimagic squares is approximately the same as for regular magic squares.
Then we can apply the following approximation formula:
 N(m) ≈ N~(m) = C~(m) · E(m) with the approximation C~(m) = 0.185 · (m - 1)

Bimagic squares

 Number N of bimagic squares Order What we estimated 2005 What we know 2014 7 4 · 10-4 N = 0 8 2 · 1007 N = 26,158,848 9 1 · 1022 N >> 1 10 2 · 1040 N >> 1 11 3 · 1065 N >> 1 12 4 · 1096 N >> 1
N >> 1 means that there are much more than one bimagic square of that order.
Christian Boyer, Pan Fengchu and I did exhaustive computer searches which showed that there are no bimagic squares of order 7. But we haven't got an analytic proof. I assume that there is no convincing mathematical reason for the nonexistence of these squares. It is just a matter of statistics. The probability for the existence of an order-7 bimagic square is very low.

Strictly pandiagonal bimagic squares

On his above mentioned web site Christian Boyer askes:
'Who will be the first to construct a normal pandiagonal bimagic square, where all broken diagonals are bimagic.'
I would like to call those squares strictly pandiagonal bimagic. What smallest order can we expect for such a square?
Here are the E-numbers for strictly pandiagonal bimagic squares (4m - 4 unforced magic lines):
Ep,b(12) ≈ 1.8 · 10-31
Ep,b(13) ≈ 3.7 · 10-15
Ep,b(14) ≈ 3.4 · 1008
Ep,b(15) ≈ 3.3 · 1037
Ep,b(16) ≈ 2.3 · 1071

Even if the probability for symmetrical squares may be a little greater and even if the C-factor is large, it is very unlikely that strictly pandiagonal bimagic squares of orders ≤ 13 exist. The probability increases enormously for orders ≥ 14. But we know pandiagonal magic squares of order 14 do not exist.
 I assume: the smallest strictly pandiagonal bimagic squares are of order 15.

Trimagic squares

 Number N of trimagic squares Order What we estimate What we know 11 6 · 10-34 N = 0 12 1 · 10-7 N >> 1 13 2 · 10-1 ?

Together with Christian Boyer I proved in 2002 that there are no trimagic squares of orders m ≤ 11.
Also according to statistics the bimagic squares of order 11 would be very unlikely.
Bimagic squares of order 13 may exist. That's an open question.

 What is wrong with the trimagic order-12 estimate?
I found trimagic squares of order 12 in 2002. There are plenty of them. But statistics say they shouldn't exist.
Symmetrical magic order-12 series consist of 6 pairs of complementary numbers (the sum of each pair is 122+1 = 145). Among the 45,828,982,764 bimagic series there are exactly 36,650 symmetrical ones. Hence only about 1/1250000 of all bimagic order-12 series are symmetrical.
Note that each of the symmetrical bimagic series is trimagic (easy to prove). Thus there are exactly 36,650 symmetrical trimagic series of order 12. All in all we have 2,226,896 trimagic series and about 1/61 of them are symmetrical.

Let's estimate the number of symmetrical trimagic order-12 squares.
We consider squares that are symmetrical with respect to a vertical axis.
First we have to count the symmetrical number squares of order 12:
We have 72 places for 72 complementary pairs. Additionally we can interchange both numbers of any pair: 72! · 272 ≈ 2.89 · 10125
The number of combinations of 6 complementary pairs selected from 72 ones is: 72! / (66! · 6!) = 156,238,908
The probability for the first row to be trimagic: prow = 36,650 / 156,238,908 ≈ 2.35 · 10-4
There is a significantly different probability for the columns. Each column consists of 12 numbers with no complementary pairs. There are 580,762 such trimagic order-12 series. How many combinations of 12 numbers without any complementary pair do exist?
144 · 142 · 140 · ... · 122 / 12! = (72! · 212) / (60! · 12!) ≈ 6.29 · 1016
The probability for the first column to be trimagic: pcol ≈ 580,762 / 6.29 · 1016 ≈ 9.23 · 10-12
Due to symmetry the 6 columns on the right are magic if the 6 columns on the left are magic. We only have to check 6 columns and one diagonal (the second is magic if the first is). The symmetrical trimagic E-number:
 Es,t(12) ≈ 2.89 · 10125 · prow11 · pcol7 / 8 ≈ 2.4 · 107
What a great result! There should exist some Millions of symmetrical trimagic order-12 squares, and apparently they do.
I wonder if there is a single trimagic order-12 square where complementary pairs are not in the same row or column.