What is the relationship between drag and sink rate?
total drag (mostly induced drag at high AOA) exceeds thrust available and the airplane develops a sink to try to regain equilibrium.


That is exactly right. At best L/D, induced drag is half of total drag and at lower speeds it becomes dominant, growing with the inverse of airspeed squared.


  Some people have said that weight exceeds lift, but I'm pretty sure this isn't the case.


You did well not to listen to those people. If the aircraft doesn't accelerate downwards, lift still equals weight. At a constant sink rate the aircraft does not accelerate.

As you say, the aircraft is in an equilibrium. Because you choose to add less power than what was necessary for level flight, the aircraft behaved similar to a glider. Allow me to use a glider for my explanation. Below is one with the dominant forces added as arrows in level flight. Clearly, there is nothing to compensate drag (red):



This airplane is not in equilibrium and does what you did in order to slow down: It flies level while drag is exceeding thrust (which is per definition zero in a glider).

So what to do? Just like when flying a turn, the pilot will now tilt the lift vector. That is the biggest force at his command, and he does so by pitching down. Now the airplane is in a glide, the direction of flight is slightly down and the lift vector (which is per definition perpendicular to the direction of the airspeed, which again is equal to the direction of flight if there is no wind) is equally tilted forward. This is what the next sketch shows:



Now we have a component of lift pointing forward, and that component is equal and opposite to drag. The rest of the lift is needed to compensate the weight, so the lift force is a tiny amount larger than weight. The sink rate is speed times the tilt angle, which in turn is what is needed to have the sine of lift $L$ equal drag $D$. Mathematically speaking, this is:
$$D = L\cdot sin(\gamma)\;\;\;\text{and}\;\;\;\gamma = arcsin\left(\frac{D}{L}\right)$$
with $\gamma$ as the angle of the direction of flight relative to the horizon.
The sink rate $v_z$ is $$v_z = v\cdot sin(\gamma) = v\cdot\frac{D}{L}$$
That is the direct answer to your question: The sink rate is the product of speed and the inverse of the lift-to-drag ratio. At small angles the ratio is approximately equal to its sine when expressed in radians, so I left the trigonometry out of the last equation.