The assumption you made is intuitive and correct, however this is practically already accomplished.
The resistance is analogous to water (or air) flow in a pipe, and the resistance from the walls of the pipe are effectively our predominant resistance. However, when you are measuring resistance in a pipe, the end effect is a drop in pressure from one end to the other, which is similar to voltage potential. To make electricity flow from one end to another, you have to have a voltage potential. Well, flow through a pipe is no different, and you require a pressure potential.
If you were to measure the pressure of the air at the exit of the compressor, it would be 90 PSI, 2.6 SCFM. If you measured it at the inlet of the piping to your tool, it might be 89.99 PSI, 2.6 SCFM.
So, at THAT point, you need to actually increase the speed of your compressor to put more air through at the same pressure.
If you care to understand that concept:
To compress air, we have to do something called Pressure-Volume work (PV work). In this case, you'll be doing PV work onto the system, or the air you want to compress.
This looks like:
Note that normally there's a negative sign in front of the integral, but in this case we're doing work on the system, so the sign is positive indicating that work is being applied to the air we're talking about.
if you change the $V$ to $V/second$ or a flow rate (such as 2.6 SCFM), then work (joules) becomes power (joules/second or watts). That is effectively limited by the motor on your compressor, or any mechanical limitations of a compressor (such as a max volume per cycle, and a maximum cycles per unit time etc)