What is intuitively the Hodge dual of a $p$-form? Carroll in his textbook "Spacetime and geometry" defines the Hodge dual of a $p$-form $A$ on an $n$-dimensional manifold as $$(\star A)_{\mu_1...\mu_{n-p}}=\dfrac{1}{p!}\epsilon^{\nu_1...\nu_p} _{\ \ \ \ \ \ \ \ \ \ \ \mu_1...\mu_{n-p}}A_{\nu_1...\nu_p}.$$
Although this is pretty straightforward, I find that I lack any intuitive/geometrical understanding of this. What does the dual actually mean?
 A: The $\epsilon$ symbol is (proportional to) the full-rank "volume form", which you can think of as a little full-dimensional box sitting at every point on your manifold. (Note that this is a different intuitive picture from the equally valid one used in knzhou's answer.) Each index corresponds to a different direction, or to one edge of the box coming out of one particular corner. Contracting the $\epsilon$ symbol with a $p$-form defined on a $p$-dimensional submanifold "collapses" the $p$ edges of the box that run parallel to the submanifold, which leaves a lower-dimensional box spanned by the $n-p$ remaining edges, which are clearly all perpendicular to the $p$ edges that got collapsed. (Note that by using the notion of "perpendicular", you're implicitly using the metric, which is hidden inside the mixed signature of the $\epsilon$ symbol in the equation you wrote above.) That motivates why the Hodge dual of a $p$-form is an $n-p$ form, and also why it's the local, point-by-point "orthogonal complement" of the original form, as knzhou explained.
A: I don't know if this is too basic for you, but think of freshman electromagnetism, where in order to define flux we associate an area vector with a small area on a surface. The area would more naturally be described by two vectors, $u$ and $v$, lying in its plane, such that the parallelogram they span has the area we're talking about. When we represent this area instead by a single vector $a$ perpendicular to the surface, we're taking the Hodge dual of the rank-2 tensor $u^av^b$. This works in three dimensions because 1+2=3, so the Hodge star converts a rank-2 tensor to a rank-1 tensor.
A: There's a simple way to visualize differential forms that helps here. 
One of the most important properties of differential $p$-forms is that they are tensors that can naturally be integrated over a $p$-dimensional submanifold. The simplest example is a one-form, which can be integrated over a one-dimensional curve $C$,
$$\int_C A = \int A_\mu dx^\mu.$$
Unlike the integration of a vector, 
$$\int \mathbf{F} \cdot d\mathbf{x}$$
this integral requires no metric to define. Similarly, a two-form can be naturally integrated over a surface, for example
$$\int_S F = \int F_{\mu\nu} dx^\mu dx^\nu$$
could be interpreted as a magnetic flux if the surface is spatial.
This leads directly to the geometric interpretation of differential forms. The familiar visualization of a one-form is a set of hypersurfaces, i.e. submanifolds of codimension one. Similarly, two-forms may be interpreted as submanifolds of codimension two, and so on. In three dimensions, a one-form is a bunch of planes, a two-form is a bunch of lines, and a three-form is a bunch of points. 
You can easily count the number of planes a curve goes through, the number of lines a surface intersects, and the number of points a volume contains. These are the geometric interpretations of integrals of one-forms, two-forms, and three-forms over curves, surfaces, and volumes. (The main thing missing from this picture is the sign; you should also think of the surfaces, etc. as coming with orientations, though that's harder to draw.)
The wedge product can be interpreted as taking the geometric intersection of the differential forms involved. For example, the wedge product of two one-forms in 3D is a set of intersections of surfaces, which is a set of curves, i.e. a two-form.
The interpretation of the Hodge dual, which is explicitly defined via the metric, is that it is the "orthogonal complement" of a differential form. The dual of a one-form in 3D is a set of curves perpendicular to the planes of the original one-form, so that the local densities of surfaces in both are proportional. 

For completeness, here are a few more facts.
First, the exterior derivative $d$ corresponds to taking the boundaries of the surfaces associated with a differential $p$-form. That is why $d^2 = 0$ is just the usual "boundary of a boundary is zero" principle, and why a closed differential form is annihilated by $d$; its visualization is in terms of closed surfaces. 
Second, one can also interpret the contraction of a $p$-form with a $(p, 0)$ antisymmetric tensor geometrically. A $(p, 0)$ antisymmetric tensor is called a $p$-blade and is constructed as the antisymmetric product of $p$ vectors. It can be interpreted as the $p$-dimensional volume element corresponding to the parallelepiped spanned by the vectors; for example the $dx^\mu dx^\nu$ above is a $2$-blade that is a small square in Cartesian coordinates. 
Third, my interpretation is not the most common one. Most books treat $p$-forms in the exact same way as $p$-blades, i.e. as small volume elements. This is the point of view advanced in the three other answers to this question. It's a little simpler, but to use it you have to explicitly use the metric to convert forms to blades. My interpretation just generalizes the intuition we are usually given for $1$-forms as families of hypersurfaces, without needing to use the metric to modify the object.
A: If you think of a blade as a subspace together with a real number, then the Hodge dual is the dual subspace together with the same real number.
