Greybeard has it basically correct. You're looking at an overturning moment though, not torque. It's force x offset to point of overturning, forces passing through the overturning point do not contribute to overturning. Meaning you could apply downward force to the front face of the tank, and nothing happens until the tank or stand fail from the stress being applied.
Assuming the front of the tank is the point of rotation, you'd have:
1000# acting down, at 0.75' horizontal offset from the front of the tank
Y# acting down, 0.25' from the tank - on the 3" lip on the stand, assuming a kid is hanging from the stand directly. 48.5" stand leaves 0.25' on each side of the tank.
To solve for Y:
1000# x 0.75' = Y# x 0.25' - right side of the equation is trying to rotate the tank, left is resisting. To overturn, you need "Y" to be large enough for the right side to be bigger than the left.
Y > 3000# to rotate the tank. (750 #-FT/0.25 FT)
Extend the moment arm to 16" (1.3') for hanging on a cabinet door as Greybeard suggests, and Y > 576#
Let's go a bit further, and assume a 50# toddler. What does the moment arm need to be to flip the tank?
(1000# x 0.75')/50# = 15'.
A 200# adult, body weight only? 3.75'
If you tried to flip the stand in a fit of rage?
750#-ft/1.5ft => Needing to apply over 500# of force.
It all seems very implausible to me that a child, or even adult, would flip a full 90g tank by hanging on the stand or the tank.
Push down on the front of the stand with your body weight, I'm sure you already know without math the tank won't go anywhere. Maybe a show of physics in action will make the wife feel better.
Edit: Post above me went up while I posting this, we have some different answers, I'm confident in mine, but our conclusions are the same.