Will increased weight slow a pendulum? When a pendulum is swinging, if you could instantly add significant weight to the bob without altering it in any other way, will that weight increase drag and cause the pendulum to move to equilibrium MORE QUICKLY?
 A: Lots of aspects to this question.
Mass per se does not increase drag - volume of the bob of the pendulum might. The increase in mass (inertia) makes the stored energy of the bob larger (thus - longer time for motion to decay); but the projected area (in the direction of the motion of the bob) will presumably also increase, which result in greater drag.
There is also a minor question of the period of the pendulum; for a pendulum with a rod of finite mass, adding mass to the bob will increase the effective length and thus slow the period of the pendulum which results in it taking longer to come to equilibrium.
All this can be expressed in terms of equations - but how you use them depends on the exact circumstances of your situation.
Period of oscillation:
$$T = 2\pi\sqrt{\frac{I}{m\;g\;\ell}}$$
Where $I$ is the moment of inertia about the pivot, $\ell$ is the distance from the center of mass to the pivot. Of course $I$ depends in part on the rod, and in part on the additional mass.
Damping: for lightly damped systems, the amplitude envelope follows the form $e^{-\gamma t}$ where $\gamma = \frac{c}{2m}$. If $c$ (drag coefficient - which is a function of shape of the bob) increases more slowly than $m$ (mass of the bob), then oscillation will take longer to damp down. In fact, drag can be quadratic with velocity in which case the usual analysis for damped SHO doesn't quite work, but the underlying principle is the same.
"It depends".
A: I don't know what you mean by "drag" exactly, but if you've got friction forces at play, then the increased mass can certainly increase both inertia and restoring force, making the friction forces less important by comparison (though not decreasing them per se). 
However in general if friction is not an issue, the gravitational force that tries to bring the pendulum back to the center is increased by the exact same factor as the inertia of the pendulum -- its resistance to accelerate -- and its speed will stay the same. This is why, as @Hennes points out, the period of a pendulum is purely determined by the length of the string involved and the local gravitational acceleration.
It is essentially a point of dimensional analysis: you have something with units $[[g]] = \text m/\text s^2$, and something with units $[[L]] = \text m$, and you can't cancel out the unit of $[[m]] = \text{kg}$, so it must have no bearing on the physical quantity $[[T]] = \text s$.
