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my experience in audio is that for IIR filters, there is usually a cascaded Second-order Section (SOS) for every little feature (a bump or an edge in the frequency response deliberately placed there by the user). often just a single second-order IIR is used for many little jobs where some frequency is given a little boost or cut. we call those IIR filters a Parametric Equalizer.
however, for a fixed and sharp feature (like a brickwall of some sort), then the order of the IIR filter will be related to the sharpeness of that feature. for a brickwall, there is a 6 dB per octave of steepness in the wall for every integer order of the filter.
for FIR, Fat's answer is good. the order of the filter is 1 less than the number of taps of the FIR. the minimum order again is related to the sharpness of some particular feature in the frequency response. there are a couple of heuristic formula for initial guessing at what order you will need. i will try to express them in a way to compare apples-to-apples.
using a Kaiser-windowed FIR design the filter order (one less than the number of taps) is about:
$$ N \ge \frac{(A-7.95)F_\text{s}}{14.357 \cdot \Delta f} $$
where $A$ is the amount of attenuation in dB of the brickwall, $\Delta f$ is the frequency width of the transition between the nominally un-attenuated and attenuated portions of the frequency response, and $F_\text{s}$ is the sample rate.
using Parks-McClellan will reduce $N$ a bit:
$$ N \ge \frac{(\tfrac12 A + \frac{4.343}{r_\text{p}} - 29.7)F_\text{s}}{14.602 \cdot \Delta f} $$
where the new symbol $r_\text{p}$ is the passband ripple in dB.
this appears to be twice as good as i was expecting. so i am not sure i did this right, but i was being very careful with the equations from O&S about the P-McC alg.
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