Why is the pressure on the ground equal to the weight of the atmospheric column on it? On the table top, the coins are stacked together, and the pressure on the table top is equal to the weight of the coin column, because the coins are in contact with each other. The atmospheric molecules are not in contact with each other. There is a large space between them. How can the pressure on the ground equal the weight of the atmospheric column? Is there any experimental proof?
So, what has happened is that you have over simplified your explanation for the stack of coins, and now you're having trouble simplifying the atmosphere in the same way. The actual explanation is the interplay of two principles so similar that they are regularly confused by undergraduates.
The first question is, why is any given coin not accelerating upward? The answer is given by Newton's first law, it is in a state of force balance, where the up-force is exactly the same as the down-force. The second question is, what does Newton's third law now state? It states that the up-force on any one coin must be compensated by an equal down-force on the coin underneath it.
That direct physical contact is totally irrelevant can be seen with magnets. If you have a magnet shaped like a washer with its north-south axis along the axis of rotational symmetry, then many of them together can be stuck together into a cylinder with a hole cut out of the middle. But if you stick a dowel rod through the middle and orient two magnets opposite, they will repel. If the dowel rod sticks vertically out of a table, then one of the magnets will float in mid-air, and it will not touch the lower magnet. But Newton's third law still requires the lower magnet to feel the same magnetic force that the upper magnet does, just opposite: and Newton's first law still requires that magnetic force to oppose the force of gravity for the floating magnet, otherwise it would be accelerating.
Air molecules in the atmosphere are constantly bouncing every single way, so force balance is not a perfect description. However, even though it doesn't hold for individual air molecules, it holds on average for a large volume containing many of them: in this case what we are looking at is not an unchanging individual momentum but a total momentum now also able to follow by exchanging particles with other nearby volumes. Still the total momentum in the volume must stay constant, and then conservation of total momentum requires the lower volume that pushes up to balance the upper volume.
On the table top, the coins are stacked together, and the pressure on the table top is equal to the weight of the coin column, because the coins are in contact with each other. The atmospheric molecules are not in contact with each other.
Even the coins are not in a total contact with each other. If you can magnify the space between each two successive coins, you can see that too many electron shells of one coin tend to repel the electron shells of the other keeping a very small void between the last atomic shells of each coin. The difference is just, in atmospheric atoms/molecules, this void is of greater magnitude.
The relevant forces (gravitational weights) are conveyed through these intense electric fields produced between the coins or atmosphere atoms regardless of whether they are in complete contact. Remember that if you bring two $S$ poles of two magnets close to each other, your hands feel force, though there is still a gap between the two magnets and they are not in contact yet.
Pressure is proportional to both the number of collisions per unit time and the speed of the collisions. Molecules at the top have more potential energy than those at the bottom. This creates a redistribution of either velocities or density, because as molecules move and collide the ones at the top will gain speed and transfer more momentum to the ones on the bottom. For an ideal gas $P=\rho kT/M$ (with M the molecular mass), so a change in pressure will necessarily be associated with a change in density, temperature or both.