Question: I work with low-power DC voltage regulators. I am already aware of the formula to calculate the size of smoothing capacitor(s). This can be an iterative process of testing one size with a scope and then using a larger size or adding more until the scope shows acceptable (very low) levels of ripple and noise.

Besides the cost of the capacitors, is there any tradeoff to rounding up (a lot) and just using a very large capacitor(s) rather than trying to calibrate the sizing to "just enough" but not more than that?

Answer: 
A larger than minimum smoothing capacitor on the output of a transformer and rectifier will give you lower ripple, which is a plus. It's a small plus however, as even doubling the size of the capacitor will only (roughly) halve the ripple. Anything downstream of a large capacitor will need to have significant Power Supply Rejection Ratio (PSRR) to cope with the ripple. There are cheaper ways of improving this by a factor of two than doubling the size of the Big Filtering Capacitor (BFC).

The downside to a larger BFC is that it will draw larger, shorter current pulses from the input transformer and rectifier.

This can cause a number of problems, though most are small, or can be mitigated.

a) Higher electromagnetic interference generation, due to larger current pulses, and higher currents being switched off in the diodes.

b) Slightly hotter diodes and transformer, due to larger RMS current.

c) Poorer input power factor.

A sniff of inductance somewhere in the supply (AC input, transformer leakage inductance, post transformer or post diode) will reduce the magnitude and extend the length of the rectifier pulses, improving all of the above.
Larger capacitors also have more parasitics (eg equiv. series resistance and inductance.) This is what "slows them down" so to speak.