Let's say I took a tiny metal sphere that when put under water has the surface area to allow, at any point in time, to be surrounded by up to 1000 water molecules. Now lets say we put this sphere first in shallow water and then in the Mariana Trench. Obviously, the sphere would feel much more pressure in the deep water! But why is that? Let's look at the formula's:
For the pressure from the 1000 water moleculse:$$P=1000*\frac{F_{molecule(H2O)}}{A_{sphere}}$$
Assuming the collisions of the water molecules with the sphere are perfectly elastic and are always in the same direction and happen in the same period of time: $$P=1000*\frac{2m_{molecule(H2O)}*v_{molecule(H2O)}}{A_{sphere}*\Delta t}$$
So the only variable here is the velocity of the water molecules. But we know that deep water is colder than shallow water so the kinetic energy, and thus velocity, of the water molecules in the mariana trench is lower and so it doesn't make much sense that the pressure would be higher.
P.S. To be explicit. My logic here is that the only things capable of DOING the force (to cause pressure) are the water molecules directly in contact with the sphere. That is where an exchange of energy would be happening.
Answer
The problem is that you're modeling the liquid like an ideal gas, whose molecules independently bounce off the ball, but liquids are characterized by strong interactions at short distances.
A better (but still inaccurate) model would be to treat the liquid like a solid locally, i.e. imagine each of the liquid molecules connected in a chain by springs. An increase in pressure means that the springs are compressed more and more, so they push outward onto your object more and more.
In terms of your variables, we should have $F \sim k \Delta x$, not $F \sim 2mv/\Delta t$. In this model, pressure can be transmitted from molecules far away, just like tension is transmitted through a rope.
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