Each object in the Universe has a huge potential energy, because you can drop it into a black hole, accelerating it to nearly the speed of light.
Is this potential energy already included into Einstein's $E=mc^2$ equation, or does it come on top of that?
$E=mc^2$ for a stationary particle in special relativity, where there are no black holes (and no gravity). So black holes (and gravity more generally) have nothing to do with this $E$.
You could say that the energy of a small test-body in the vicinity of a spherically symmetric black hole in general relativity in the case when the test-body has no velocity compared to the center of the gravitational field can be written as:
$$E=mc^2{\sqrt{1-\frac{2GM}{rc^2}}}.$$
So if you drop an object from rest at infinity so that if falls down and crashes into standstill on the surface of a planet the energy lost as heat/radiation at impact can be written as:
$$\Delta E=mc^2-mc^2\sqrt{1-\frac{2GM}{rc^2}}$$
Answer: You could say that the potential energy is not baked in the expression $E=mc^2$ but it is baked into $E=mc^2\sqrt{1-\frac{2GM}{rc^2}}$.
