As everyone has said, your answer depends on how fast it was going, and over what time or distance the thing took to stop. Or more precisely, the stopping distance for the shockwatch (as the part to which it is fixed may have had some spring in it).
We assume that the pallet was dropped cleanly, rather than sliding off something, or tipping over. We also assume that the pallet came to a damped (squishy) stop, rather stopping on a trampoline or springy floor.
Given all that, (plus a few more unstated but less important assumptions) the g force depends on the velocity of the shockwatch at contact, and the distance over which it came to rest. And the velocity on contact depends on the height from which it was dropped.
The velocity V attained by a free-falling thing is given by V squared = U squared + 2 * g * H .....(1)
where g = 9.807 metres/sec/sec; V is metres/sec; H is height in metres and U is the initial velocity, in our case zero.
The same formula describes the stopping process, except the deceleration comes from your shockwatch, and the height is the stopping distance. So v squared = V squared + 2 * G * g * h ....(2)
where v is the final velocity (zero), V comes from above (the initial velocity on impact), G is the shockwatch (and negative); h is the stopping distance.
Now if you substitute V squared from (1) into (2) (because they are the same thing) and simplify, you get
H = G * h. That is, drop height = shockmeter * stopping distance
and when you think about it, this simply is a definition of G.
So stopping distance 1cm; Drop height 5 cm for 5G; 25 cm for 25G
Stopping distance 5cm; Drop height 25 cm for 5G; 75 cm for 25G.
Doesn't take much to generate Gs.
A hell of a lot depends on where the shockwatch is placed. If it is on the top of a deformable crate, but that crate has a machine at the bottom with only the pallet between it and the concrete to deform in taking the shock, then the machine sees more G than the shockwatch.
ET