My thought process here is the following: The inside of an (eternal) black hole carries four (Schwarzschild) coordinates t, r, theta, phi – r now being timelike and confined to the interval (0, 2M) and t now being spacelike and being any real number. That is, depending on when (at what time t) you cross the event horizon, you end up at a different point in space. The singularity at r=0 is then a point in your future which, like your own death, you cannot actually see but which you will nevertheless hit in finite proper time.

So in this sense I'd say the volume is very finite (if we disregard the (trivially unbounded) spacelike coordinate t which, as mentioned before, simply corresponds to the time of entering the BH).

×) Counterexample: Consider the manifold M := R³\B, where B is the closed unit ball, equipped with the standard Euclidean metric. This manifold is certainly not Cauchy-complete and we can reach the singularity at r=1 in finite time. Now, if we had to define the dimension of the singularity, what dimension n should it have? n=2 (a sphere)? Maybe. At least we could extend M by the unit sphere to make it complete. But could the singularity also be a point (i.e. n=1)? Yes, certainly. By diffeomorphism invariance, we could simply find new coordinates and map R³\B to R³\{0}, so the singularity would suddenly become a point. So, as you can see, interpreting the singularity as a point or set of points that have a topological dimension doesn't work.

>A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point.

In Euclidean geometry,

A cube is 3 dimensions.

A plane is 2 dimensions.

A line is 1 dimension.

A point is ...

> In this diagram the singularity is a line in spacetime i.e. a one dimensional object in spacetime.

is wrong or at least very misleading – the answer does (correctly) say that asking for the "dimensionality of a singularity […] is a meaningless question because the spacetime geometry is undefined at a singularity".