Does the hyperdeterminant calculate a quantity akin to the volume of a parallelepiped? If $M$ is an $n \times n$ matrix, $|\det(M)|$ is the volume of the $n$-dimensional
parallelepiped spanned by the column vectors of $M$.

          


          

(Image from Wikipedia's Determinant article.)


In the 19th-century, Cayley defined the hyperdeterminant of a hypermatrix $H$.
(A hypermatrix 
can be viewed as a representation of a
tensor.)

My question is:

Q.
  Does the hyperdeterminant have a geometric interpretation somehow analogous
  to the parallelepiped-volume interpretation of the determinant?

Perhaps an answer resides in the
book by Gelfand, Kapranov, and Zelevinsky entitled Discriminants, Resultants and Multidimensional Determinants (Birkhäuser, Boston, 1994;
MAA link),
which I have yet to
examine.
 A: At least, in the classical case treated by Cayley, the $2{\times}2{\times}2$ hyperdeterminant, there is an interpretation in terms of volumes generalizing the classical determinant case.  It goes like this:
First, recall that, when two vector spaces $U$ and $V$ have the same dimension, say, $r$, there is a polynomial map of degree $r$
$$
\mathbf{det}: U\otimes V \to \Lambda^r(U)\otimes\Lambda^r(V)
$$
defined, relative to any basis $u_i$ of $U$ and $v_j$ of $V$, by 
$$
\mathbf{det}( a^{ij} u_i\otimes v_j) = \det(a^{ij})
\,(u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes (v_1\wedge v_2\wedge\cdots\wedge v_r)
$$
where $\det$ is the usual determinant of a matrix.  It is easy to see that this definition of $\mathbf{det}$ is independent of the choice of basis $u_i$ of $U$ and $v_\rho$ of $V$.  Note that, since $\Lambda^r(U)$ and $\Lambda^r(V)$ are the top exterior powers of $U$ and $V$, their elements represent 'volume elements' (sometimes called 'mass elements') in $U$ and $V$ respectively, so a tensor $\alpha\in U\otimes V$ gives rise, via $\mathbf{det}(\alpha)$ a way to 'multiply' volume elements in the two vector spaces.  
(Of course, in the usual linear algebra interpretation, we consider $\alpha\in U^*\otimes V$ as a linear map from $U$ to $V$, and then 
$$
\mathbf{det}(\alpha)\in \Lambda^r(U^*)\otimes \Lambda^r(V)\simeq
\Lambda^r(V)\otimes \Lambda^r(U)^*\simeq\  "\Lambda^r(V)/ \Lambda^r(U)"
$$
can be thought of as a ratio of volume forms on the two vector spaces.  Even more specially, we can take $U=V^*$, and then, because $\Lambda^r(V)\otimes \Lambda^r(V)^*$ is canonically isomorphic to $\mathbb{R}$, $\mathbf{det}(\alpha)$ becomes just a scalar.)
Now, when you have three vector spaces $U$, $V$, and $W$ of the same dimension $r$, you can expand a tensor $\alpha\in U\otimes V\otimes W$ in terms of a basis $u_i$ of $U$, $v_j$ of $V$, and $w_k$ of $W$ as
$$
\alpha = a^{ijk}\,u_i\otimes v_j\otimes w_k
$$
and you can just regard the $w_k$ as 'indeterminates' and define
$$
\mathbf{det}_{UV}(\alpha) = \det(a^{ijk}w_k)\otimes
(u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes(v_1\wedge v_2\wedge\cdots\wedge v_r),
$$
where, now,
$$
\mathbf{det}_{UV}(\alpha) \in S^r(W)\otimes \Lambda^r(U) \otimes \Lambda^r(V).
$$
To go further, you need to look at particular values of $r$.  What Cayley did was consider the fact that, for quadratic forms in $r$ variables, there is a well-defined discriminant mapping $\mathrm{discr}_W:S^2(W)\to S^2(\Lambda^r(W))$ that is a homogeneous polynomial of degree $r$.  It is defined by 
$$
\mathrm{discr}_W(q^{ij}w_iw_j) = \det(q^{ij})\,
(w_1\wedge w_2\wedge\cdots\wedge w_r)^2. 
$$
In particular, when $r=2$, we can compose to get an element
$$
\mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha))
\in S^2(\Lambda^2(U))\otimes S^2(\Lambda^2(V)) \otimes S^2(\Lambda^2(W)),
$$
which is a polynomial of degree $4$ in the coefficients of $\alpha$.
This is (the negative of) Cayley's hyperdeterminant in the $2{\times}2{\times}2$
case. Note that it is, indeed, expressed in terms of volume forms on the three vector spaces $U$, $V$, and $W$.  It's just that it is now the product of squares of volume forms.  By the way, it is not hard to show that, if we had, instead, computed $\mathbf{hdet}_{VWU}(\alpha)$, we would have got the same result.  
I think that, at least in the $r=2$ case, this is probably the best interpretation of the hyperdeterminant in terms of volumes.
In the case when $(\dim U, \dim V, \dim W) = (2,2,s)$ where $s > 2$, this gives an expression
$$
\mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha))
\in S^s(\Lambda^2(U))\otimes S^s(\Lambda^2(V)) \otimes S^2(\Lambda^2(W))
$$
of degree $2s$, but, of course, this vanishes identically when $s > 4$.
When you go to higher values of $r$, it's not so clear.  For example, when $r=3$ (again, with all vector spaces of the same dimension $r$), there are the two Aronhold relative invariants:  $Q^4: S^3(W)\to S^4(\Lambda^3W)$ (of degree $4$) and $Q^6:S^3(W)\to S^6(\Lambda^3W)$ (of degree $6$), and so we can define two expressions
$$
\mathbf{Q}^4_{UVW}(\alpha) = Q^4_W(\mathbf{det}_{UV}(\alpha))
\in S^4(\Lambda^3(U))\otimes S^4(\Lambda^3(V)) \otimes S^4(\Lambda^3(W))
$$
of degree $12$ and 
$$
\mathbf{Q}^6_{UVW}(\alpha) = Q^6_W(\mathbf{det}_{UV}(\alpha))
\in S^6(\Lambda^3(U))\otimes S^6(\Lambda^3(V)) \otimes S^6(\Lambda^3(W))
$$
of degree $18$.  Thus, these clearly relate powers of volume elements of the underlying vector spaces.  According to the Wikipedia page, though, the hyperdeterminant of $\alpha$ in this case must have degree $36$; presumably it is a polynomial in the above two invariants.
A: This might help, although I can only visualize it in $3$D. In your opening sentence you write:

spanned by the column vectors of M.

The $n\times n\times n$ hyper-matrix (assuming it is non-degenerate) has columns that are linearly independent in $\mathbb{R^3}$. If we place these in $\mathbb{R^3}$ and join them up again (similar as to a parallelgram), we obtain a parallelapiped.
The hyper-determinant therefore represents the hyper-volume of the associated parallelapipeds structure created from the hyper-matrix.
EDIT
Or, consider a parallelapiped with a basis a,b,c. The vertices are formed by the linear combinations a+b,a+c,b+c,a+b+c. With a $3\times3\times3$ hypermatrix, read top to bottom, we have a collection of $9$ vectors, and the vertices of the resulting shapes are the linear combinations of these vectors.
