... does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?

I'm not completion sure what distinction you're drawing between concentration and density, but I will assume that what you mean by the former is the details of the matter distribution, e.g., whether it's concentrated at the center, spread throughout, or whatnot.

For a spherically symmetric isolated body of mass $M$, it is completely irrelevant. The reason is Birkhoff's theorem: outside the gravitating body, spacetime geometry is necessarily Schwarzschild. This is the general-relativistic analogue of Newton's shell theorem. Therefore, it doesn't matter whether the (radial) distribution is uniform, concentrated at the center, or some kind of shell, or anything else: once it is compact enough that its outer surface gets to the Schwarzschild radius $2GM/c^2$ or below, it is fully enclosed by an event horizon, and is therefore a black hole.

So under those assumptions, the answers to your two questions are 'yes' and 'no', respectively, although you might want to be careful about how you define 'volume' when comparing overall densities.

What happens if we get rid of the assumption of spherical symmetry is a bit more complicated. If we're in an asymptotically flat universe, then we can think of a black hole as all the event from which an ideal light ray fails to escape to infinity, and the boundary would be the event horizon; more generally, we might have to be more careful about how we define 'inside' and 'outside'. Note that this makes the event horizon depend on the future, i.e. it depends on what light rays escape or don't escape even if you wait an arbitrarily long time for them. Hence, in a dynamic situation (such as a collapse to a black hole), where the location of the event horizon depends on not only on the past and present, but also on what will fall into the black hole in the future.

This makes general statements about density quite difficult in situations that don't have some simplifying assumptions. Density is too simplistic; the general notions of a black hole and event horizon are highly non-local.

Nevertheless, there is a general result that is morally similar to the above that is very relevant to your second question: the no-hair theorem. In general relativity, any isolated black hole is fully characterized by conserved quantities at infinity (mass, angular momentum, electric charge...). That means that the details of the matter distribution inside the event horizon do not matter at all. Of course, the singularity theorems guarantee that at least under some general assumptions about the behavior of the matter, it will collapse to a singularity, but that's a separate issue.

... does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?

I'm not completion sure what distinction you're drawing between concentration and density, but I will assume that what you mean by the former is the details of the matter distribution, e.g., whether it's concentrated at the center, spread throughout, or whatnot.

For a spherically symmetric isolated body of mass $M$, it is completely irrelevant. The reason is Birkhoff's theorem: outside the gravitating body, spacetime geometry is necessarily Schwarzschild. This is the general-relativistic analogue of Newton's shell theorem. Therefore, it doesn't matter whether the (radial) distribution is uniform, concentrated at the center, or some kind of shell, or anything else: once it is compact enough that its outer surface gets to the Schwarzschild radius $2GM/c^2$ or below, it is fully enclosed by an event horizon, and is therefore a black hole.

So under those assumptions, the answers to your two questions are 'yes' and 'no', respectively, although you might want to be careful about how you define 'volume' when comparing overall densities.

What happens if we get rid of the assumption of spherical symmetry is a bit more complicated. If we're in an asymptotically flat universe, then we can think of a black hole as all the event from which an ideal light ray fails to escape to infinity, and the boundary would be the event horizon; more generally, we might have to be more careful about how we define 'inside' and 'outside'. Note that this makes the event horizon depend on the future, i.e. it depends on what light rays escape or don't escape even if you wait an arbitrarily long time for them. Hence, in a dynamic situation (such as a collapse to a black hole), where the location of the event horizon depends on not only on the past and present, but also on what will fall into the black hole in the future.

This makes general statements about density quite difficult in situations that don't have some simplifying assumptions. Density is too simplistic; the general notions of a black hole and event horizon are highly non-local.

Nevertheless, there is a general result that is morally similar to the above that is very relevant to your second question: the no-hair theorem. In general relativity, any isolated black hole is fully characterized by conserved quantities at infinity (mass, angular momentum, electric charge...). That means that the details of the matter distribution inside the event horizon do not matter at all. Of course, the singularity theorems guarantee that at least under some general assumptions about the behavior of the matter, it will collapse to a singularity, but that's a separate issue.