I would like to point out that there is a connection between the determinant of the covariance matrix of (Gaussian distributed) data points and the differential entropy of the distribution.
To put it in other words: Let's say you have a (large) set of points from which you assume it is Gaussian distributed. If you compute the determinant of the sample covariance matrix then you measure (indirectly) the differential entropy of the distribution up to constant factors and a logarithm. See, e.g, Multivariate normal distribution.
The differential entropy of a Gaussian density is defined as:
$$H[p] = \frac{k}{2}(1 + \ln(2\pi)) + \frac{1}{2} \ln \vert \Sigma \vert\;,$$
where $k$ is the dimensionality of your space, i.e., in your case $k=3$.
$\Sigma$ is positive semi-definite, which means $\vert \Sigma \vert \geq 0$.
The larger $\vert \Sigma \vert$, the more are your data points dispersed. If $\vert \Sigma \vert = 0$, it means that your data ponts do not 'occupy the whole space', meaning that they lie, e.g., on a line or a plane within $\mathbb{R}^3$. Somewhere I have read, that $\vert \Sigma \vert$ is also called generalized variance. Alexander Vigodner is right, it captures the volume of your data cloud.
Since a sample covariance matrix is defined somewhat like: $$\Sigma = \frac{1}{N-1} \sum_{i=1}^N (\vec{x}_i - \vec{\mu})(\vec{x}_i - \vec{\mu})^T\; $$ it follows, that you do not capture any information about the mean. You can verify that easily by adding some large constant vectorial shift to your data; $\vert \Sigma \vert$ should not change.
I don't want to go to much into detail, but there is also a connection to PCA. Since the eigenvalues $\lambda_1, \lambda_2, \lambda_3$ of $\Sigma$ correspond to the variances along the principal component axis of your data points, $\vert \Sigma \vert$ captures their product, because by definition the determinant of a matrix is equal to the product of its eigenvalues.
Note that the largest eigenvalue corresponds to the maximal variance w.r.t. to your data (direction given by the corresponding eigenvector, see PCA).
Best Answer
Let's review how the covariance matrix is computed in this context. Let $\mu_j$ denote the mean of the $j$th column, and let $\mu$ denote the row-vector $\mu = (\mu_1,\mu_2,\dots,\mu_d)$. Then $$ \operatorname{cov}(X) = (X - \mu 1_n)(X - \mu 1_n)^T $$ where $1_n$ denotes the column vector $(1,\dots,1)^T$ of length $n$.
With this in mind, the $i,j$ entry of the covariance matrix is given by $$ \operatorname{cov}(X)[i,j] = \sum_{k=1}^n (x_{ik} - \mu_k)(x_{jk} - \mu_k) $$ So, $\frac 1n \operatorname{cov}(X)[i,j]$ is the covariance between the $i$th feature and $j$th feature, and $\frac 1n \operatorname{cov}(X)[i,i]$ is the variance of the $i$th feature.