This is a case where you want to analyze in the time domain, not the frequency domain. In other words, you don't care directly about the rolloff frequency, but rather the time constant.
The time constant of a RC filter is simply R*C. When R is in Ohms and C in Farads, the the result is in seconds.
If a step is put into such a RC filter, it will exponentially decay towards the new input value. For example, if both input and output are at 1 and the input goes to 0 at t = 0, then the output is:
OUT = e-t/RC
This means that every RC seconds, the output gets another factor of e closer to the input. From this you can compute the time it takes to get to any particular output level.
For example, let's say you have a low pass filter made from 4.7 kΩ in series followed by 2 µF to ground. A Schmitt trigger with 20% and 80% thresholds looks at this signal and produces a digital output in response. What is the delay to a change in the input when that input is a digital signal that has been steady for a long time?
The time constant is (4.7 kΩ)(2 µF) = 9.4 ms. In the equation above, OUT needs to go to 20% for the circuit to trigger. We therefore know that -t/RC = ln(0.2) = -1.61. That means it takes 1.61 time constants to get to 20% change remaining (80% settled). 1.61(9.4 ms) = 15.1 ms, which is how long the circuit in this example will delay digital edges after long steady periods.