Well, I don't think it's that simple. We can't use a tire or balloon analogy here. In something like a tire, lower pressure supporting the same weight requires the tire to spread out. That way lower psi has more square inches of area supporting the same weight.

The air suspension on our cars is significantly more restrictive in manner of "ballooning" with pressure changes. Think of it this way - suppose I want to support 1,000 lbs on a hydraulic jack, and I can choose to do so at 6" or 12" of height. In this admittedly extreme example, there is zero "ballooning," and therefore the height is strictly based on the volume of fluid pumped under the piston. The pressure of that fluid, at our example weight, will be exactly the same at 6" or 12" because the supported weight and the piston area supporting that weight are static. We just have more fluid in there.

While the suspension on our cars has some flexibility, it is substantially restricted compared to something like a tire or balloon. The area in sq inches the air can act on to support the weight doesn't materially vary at different heights. Lower the car 2", and you have reduced the static air volume in the bag via the control valve, but the air is supporting the same weight and doing so over "mostly" the same area. In other words, it's not restrictive like a jack cylinder would be, but it's not anything like a tire or balloon that can disburse our given weight over a much larger area. The static pressure will be lower only to the extent that more surface area is available - not in total but with respect to the area supporting the vehicles weight.

Separately from that, an air suspension is rising rate. If the area is fixed, and we force the volume to decrease, the pressure will increase. Air is compressible, of course, so we have a measure of compliance. Here's the point most folks miss - if I start with 100 cu. in. of static volume, and I decrease a given amount (bump), the pressure will rise a determinable amount. We all know that. But if I started instead with only 50 cu. in. of static volume, and I decreased the volume by the same bump, then the pressure delta, the increased pressure resulting from that volume change, would be twice as much. Rising rate characteristic means each subsequent increase in compression (bump) has a much larger effect on my spring rate than had I started with a higher static volume. "Compression ratio" if you will. Our suspension volume isn't that rigid, but it is not entirely flexy/squishy. We're somewhere in between.

Again, if the air suspension had a substantial capacity to balloon out and spread that pressure delta over much more surface area, then out compressed pressure would not spike much - it would disburse over the ballooning area. We don't have that extreme flexibility, though.

The point is that lowering the suspension will have a lower static pressure, just not as much as folks might think, because the surface area remaining to hold up the same vehicle weight is not markedly greater. Same weight, not much more area to apply the pressure to support the weight = small change in the pressure, but mostly a change in the volume via the valve (my jack analogy).

Smaller static volume, compressed a given amount by a bump, will have a correspondingly greater rate increase over that compression cycle vs that with the larger beginning static volume.

Noticeably firmer ride on rough roads.