Here's a fun game to play that's in the spirit of the OP's question.

Pretend your electronics are fried (calculator included), you forgot the formula(s) for estimation, and your Mildot Master burst into flames.

Pick a target size and angular subtension at (somewhat, keep it reasonable) random.

So, for poops and giggles let's take a target size of two feet that takes up five mrad in the scope.

All I need for the entire process is my knowledge that one yard takes up 1 mrad at 1000 yards and a little good old-fashioned common horse sense. You can break the process into two separate steps. Adjust for the angle and adjust for the size.

Adjust for angular subtension: Since the target's taking up five mrad instead of one, that puts the hypothetical one-yard target that we started with at 1/5th the starting 1000-yard distance, or two hundred yards. Forget about the size difference for this step, we'll take care of that next.

Adjust for target size: Since the actual target's smaller by a factor of 2/3 than the hypothetical 1-yard target, that puts the target closer by the same factor. You know it's closer than the hypothetical because a smaller target needs to be closer to take up the same angular subtension. Two thirds of 200 yards is 133 yards.

So after the dust settles our solution is that a two-foot target that appears to be 5 mrad in height is 133 yards away.

The first numbers that came to my head resulted in a pretty precise answer. In practice you're not going to be so certain of the target size or the angular subtension, so you can relax on the significant digits and still come to a useful answer.

Bonus points: Without redoing the calculation, what would be the distance of a two-foot target that takes up .5 mrad of angular subtension?

Caution: Don't play this game while you're driving.