Not an answer, but an amusing observation: The determinant of the matrix usually (but not alwasy) has denominator a perfect square (empirically), while the numerator always seems to have a huge prime factor.
EDIT Actually, it seems that the expectation of $\log(P(n))/\log(n),$ where $P(n)$ is the largest prime factor of $n$ approaches the Golomb-Dickman constant, which is about 0.62. In view of this, the largest prime factor of the numerator is pretty much par for the course, or at least, not obviously NOT par for the course.
n=1: det = 1/1
n=2: det=7/(12^2)
n=3 det = 647/(2160^2)
n=4 det = (19 * 571)/(672000^2)
n=5 det = (179 * 179357)/(7*4233600000^2)
n=6 det = (97 * 157 * 384191938531)/(186313420339200000^2)
n=7 det = (23 * 1280587616051046200369)/(2067909047925770649600000^2)
n=8 det = (317 * 6337 * 25997 * 87403 * 511645991608091)/(365356847125734485878112256000000^2)
(and some more. Note that mysterious 7 which keeps popping up in the denominator, for n=5,9,11 )
n=9 det = (55091 * 7731550926975871647518813143593349)/(7 * 146968826339795671126721851844198400000000^2)
n=10 det = (257 * 47360083 * 530916328215423816923887043836865928533)/(15402297982638230438765209613012092908994560000000000^2)
n=11 det = (31 * 1193 * 2647 * 538580971 * 957346850101 * 71222443011485886519799225151)/(7 * 175251348661711183890804992735665222783492739007774720000000000^2))