There is a way of making sense of all integer dimensions, although I'm not sure if it will be useful to you.
Let us start with the collection of Euclidean spaces,
$$
E=\{\mathbb R^n;n\in\mathbb Z,n\geq0\},
$$
for example.
There two operations for spaces in $E$, the direct sum and the tensor product.
If we identify spaces of the same dimension, we have $\mathbb R^n\oplus\mathbb R^m=\mathbb R^{n+m}$ and $\mathbb R^n\otimes\mathbb R^m=\mathbb R^{nm}$.
These operations are associative, commutative and distributive.
These operations don't make $(E,\oplus,\otimes)$ into a ring, but that can be remedied.
One way is to formally let $n$ range over all of $\mathbb Z$ and define things the same way.
If we call this new collection of spaces $F$, then $(F,\oplus,\otimes)$ is (in an obvious way) ring-isomoprhic to $(\mathbb Z,+,\times)$.
Then objects like $\mathbb R^{-7}\in F$ don't have any other meaning than being in some kind of an extension of the set of Euclidean spaces.
I don't see any obvious way of doing geometry in an object of $F$, especially in a way that would include those volume formulas.
In this case the construction leads to a relatively trivial object (the ring of integers), but it at least gives some way of attaching a meaning to negative dimensional spaces.
That is, $\mathbb R^{-4}$ is the thing (not a vector space!) that satisfies $\mathbb R^{-4}\oplus\mathbb R^4=0$ (something like the direct sum of vector spaces), where $0=\mathbb R^0$ is the zero space.
Negative dimension cancels positive dimension, whatever that means.
(This answer was inspired by the concept of representation ring, a way of forming a ring out of vector spaces in way that is useful in representation theory.)
Sidenote: $1/\Gamma(1+n/2)=0$ for $n\in\{-2,-6,-10,-14,\dots\}$.
This would be really weird behaviour for something that resembles the geometrical idea of volume.