AFAIK a metric uniquely determines the volume element up to to sign since the volume element since a metric will determine the length of supplied vectors and angle between them, but I do not see a way to derive a metric from the volume element. The volume form must heavily constrain the the set of compatible metrics though. So is there a nice statement of what the volume element tells you about a metric that is compatible with it? If you have a volume element what is the least additional information you need to be able to derive a metric?
EDIT:
Is the minimum additional information uniform structure?
A volume element gives us a kind of notion of local scale but much weaker than a metric. We can can measure a volume, but we can't tell if it looks like a sphere, a pancake or string (that is the essence of @Rhys's answer). To get a metric we need to be able to tell that a volume is "spherical", or to compare distances without assigning an actual value to them, since if we can do that then we can use use the the volume to determine the radius of a hypershere in the limit of small volumes. So the last part of my question is really asking how to do that without implicitly specifying a metric. I believe that uniform structure does this. It specifies a set of entourages that are binary relations saying that points are within some unspecified distance from each other. An entourage determines a ball of that size around each point. To induce a metric, the uniform structure needs to be compatible with the volume element by assigning the same volume to all the balls induced by the same entourage in the limit of small spheres (delta epsilonics required to make this formal). The condition will not hold for macroscopic spheres if the uniform structure is inducing a curved geometry. There must also be some conditions for the uniform structure to be compatible with the manifold structure.
Does this make sense?
The volume form tells you very little about the metric. Let $V$ be an $n$-dimensional vector space, with volume form $v_1\wedge \ldots \wedge v_n$, where $\{v_1,\ldots,v_n\}$ are linearly independent elements of $V$ (any volume form can be written this way). Now define a metric by the condition that this be an orthonormal set; this metric gives you the above volume form. But it's far from unique, as there are many choices of $n$ vectors which give the same volume form.
The same story should apply on some small enough patch of a Riemannian manifold.