Every smooth closed orientable 4-manifold may be constructed via a handle decomposition. Before asking a couple of questions, I recall some well-known facts about handle-decompositions of 4-manifolds.

We can of course order the handles according to their index. Handles of index 0 and 1 form a connected 4-dimensional handlebody, whose boundary is a closed 3-manifold, diffeomorphic to a connected sum $\#_g(S^2\times S^1)$ of some $g$ copies of $S^2\times S^1$ (if $g=0$ we get $S^3$). Handles of index 2 are attached to some framed link $L\subset \#_g(S^2\times S^1)$. Since 3- and 4-handles also form a 1-dimensional handlebody, after the attaching of the 2-handles we must necessarily obtain a 4-manifold whose boundary is again diffeomorphic to $\#_h(S^2\times S^1)$, for some $h$ which is not necessarily equal to $g$. The new $\#_h(S^2\times S^1)$ is obtained by surgery along the framed link $L$.

Therefore, in some sense, constructing closed orientable 4-manifolds reduces to constructing framed links $L\subset \#_g(S^2\times S^1)$ that produce some $\#_h(S^2\times S^1)$ via surgery. Let us define the 3-dimensional hyperbolic volume of a handle decomposition as the Gromov norm of $(S^2\times S^1) \setminus L$ (which is in turn the sum of the hyperbolic volumes of its pieces according to geometrization, whence the name). Let us then define the 3-dimensional hyperbolic volume of a closed orientable 4-manifold as the infimum of all the hyperbolic 3-dimensional volumes among all its handle decompositions. The infimum is actually a minimum because the set of 3-dimensional hyperbolic volumes is well-ordered.

The general question is:

What can we say about the 3-dimensional hyperbolic volume of a closed 4-manifold?

A more specific one:

Which closed 4-manifolds have zero 3-dimensional hyperbolic volume?

which is equivalent to the following:

Which closed 4-manifolds admit a handle decomposition such that $\#_g(S^2\times S^1)\setminus L$ is a graph manifold?

Complex projetive plane belong belongs to this class, and also many doubles of 2-handlebodies: if you take any link $L\subset S^2\times S^1$ whose complement is a graph manifold, you can attach 2-handles to it, and then make the double of the resulting bounded 4-manifold. The resulting double has volume zero.

Finally, we have the following very specific question:

Is there a 4-manifold with positive 3-dimensional hyperbolic volume?

I would expect that most (all?) aspherical 4-manifolds have positive volume, and maybe also many simply connected ones, but I don't know the answer.

Every smooth closed orientable 4-manifold may be constructed via a handle decomposition. Before asking a couple of questions, I recall some well-known facts about handle-decompositions of 4-manifolds.

We can of course order the handles according to their index. Handles of index 0 and 1 form a connected 4-dimensional handlebody, whose boundary is a closed 3-manifold, diffeomorphic to a connected sum $#_g(S^2\times \#_g(S^2\times S^1)$ of some $g$ copies of $S^2\times S^1$ (if $g=0$ we get $S^3$). Handles of index 2 are attached to some framed link $L\subset #_g(S^2\times \#_g(S^2\times S^1)$. Since 3- and 4-handles also form a 1-dimensional handlebody, after the attaching of the 2-handles we must necessarily obtain a 4-manifold whose boundary is again diffeomorphic to $#_h(S^2\times \#_h(S^2\times S^1)$, for some $h$ which is not necessarily equal to $g$. The new $#_h(S^2\times \#_h(S^2\times S^1)$ is obtained by surgery along the framed link $L$.

Therefore, in some sense, constructing closed orientable 4-manifolds reduces to constructing framed links $L\subset #_g(S^2\times \#_g(S^2\times S^1)$ that produce some $#_h(S^2\times \#_h(S^2\times S^1)$ via surgery. Let us define the 3-dimensional hyperbolic volume of a handle decomposition as the Gromov norm of $(S^2\times S^1) \setminus L$ (which is in turn the sum of the hyperbolic volumes of its pieces according to geometrization, whence the name). Let us then define the 3-dimensional hyperbolic volume of a closed orientable 4-manifold as the infimum of all the hyperbolic 3-dimensional volumes among all its handle decompositions. The infimum is actually a minimum because the set of 3-dimensional hyperbolic volumes is well-ordered.

The general question is:

What can we say about the 3-dimensional hyperbolic volume of a closed 4-manifold?

A more specific one:

Which closed 4-manifolds have zero 3-dimensional hyperbolic volume?

which is equivalent to the following:

Which closed 4-manifolds admit a handle decomposition such that $#_g(S^2\times \#_g(S^2\times S^1)\setminus L$ is a graph manifold?

Complex projetive plane belong to this class, and also many doubles of 2-handlebodies: if you take any link $L\subset S^2\times S^1$ whose complement is a graph manifold, you can attach 2-handles to it, and then make the double of the resulting bounded 4-manifold. The resulting double has volume zero.

Finally, we have the following very specific question:

Is there a 4-manifold with positive 3-dimensional hyperbolic volume?

I would expect that most (all?) aspherical 4-manifolds have positive volume, and maybe also many simply connected ones, but I don't know the answer.

Every smooth closed orientable 4-manifold may be constructed via a handle decomposition. Before asking a couple of questions, I recall some well-known facts about handle-decompositions of 4-manifolds.

We can of course order the handles according to their index. Handles of index 0 and 1 form a connected 4-dimensional handlebody, whose boundary is a closed 3-manifold, diffeomorphic to a connected sum $#_g(S^2\times S^1)$ of some $g$ copies of $S^2\times S^1$ (if $g=0$ we get $S^3$). Handles of index 2 are attached to some framed link $L\subset #_g(S^2\times S^1)$. Since 3- and 4-handles also form a 1-dimensional handlebody, after the attaching of the 2-handles we must necessarily obtain a 4-manifold whose boundary is again diffeomorphic to $#_h(S^2\times S^1)$, for some $h$ which is not necessarily equal to $g$. The new $#_h(S^2\times S^1)$ is obtained by surgery along the framed link $L$.

Therefore, in some sense, constructing closed orientable 4-manifolds reduces to constructing framed links $L\subset #_g(S^2\times S^1)$ that produce some $#_h(S^2\times S^1)$ via surgery. Let us define the 3-dimensional hyperbolic volume of a handle decomposition as the Gromov norm of $(S^2\times S^1) \setminus L$ (which is in turn the sum of the hyperbolic volumes of its pieces according to geometrization, whence the name). Let us then define the 3-dimensional hyperbolic volume of a closed orientable 4-manifold as the infimum of all the hyperbolic 3-dimensional volumes among all its handle decompositions. The infimum is actually a minimum because the set of 3-dimensional hyperbolic volumes is well-ordered.

The general question is:

What can we say about the 3-dimensional hyperbolic volume of a closed 4-manifold?

A more specific one:

Which closed 4-manifolds have zero 3-dimensional hyperbolic volume?

which is equivalent to the following:

Which closed 4-manifolds admit a handle decomposition such that $#_g(S^2\times S^1)\setminus L$ is a graph manifold?

Complex projetive plane belong to this class, and also many doubles of 2-handlebodies: if you take any link $L\subset S^2\times S^1$ whose complement is a graph manifold, you can attach 2-handles to it, and then make the double of the resulting bounded 4-manifold. The resulting double has volume zero.

Finally, we have the following very specific question:

Is there a 4-manifold with positive 3-dimensional hyperbolic volume?

I would expect that most (all?) aspherical 4-manifolds have positive volume, and maybe also many simply connected ones, but I don't know the answer.