Square of $3/2$ dimensional fractal It is known that the following fractal (Quadratic von Koch curve, or Minkowski's sausage) has a dimension $3/2$. It means that the square $X^{2} = X\times X$ may have dimension $2 \times 3/2 = 3$, which is a natural number. Is there a good way to see that $X\times X$ is a 3-dimensional object?

 A: Preliminaries
There is a lot going on in your question:
When you say that this curve has dimension $\frac{3}{2}$, to which notion of dimension do you refer?  This matters, because your question is about the dimension of a product.  For most (metric) notions of dimension, there are results regarding the dimension of a product, but these results vary, depending on which notion of dimension you consider.  For example,
$$ \dim_H(X\times Y) \ge \dim_H(X) + \dim_H(Y),
\qquad\text{but}\qquad
\dim_B(X\times Y) \le \dim(X)+\dim(Y), $$
where $\dim_H$ and $\dim_B$ represent the Hausdorff and (upper) box dimensions, respectively.  Either of these inequalities may be strict.
Additionally, even if we make the assumption that $\dim(X\times X) = 2\dim(X) = 3$ (for some notion of dimension $\dim$), it is not clear what you mean by "seeing" this as a three-dimensional object.  This probably means something like "How do we embed $X\times X$ into three-dimensional Euclidean space?"  This is a somewhat difficult question, and the likely answer is "We cannot do this."
The Whitney Embedding Theorem
To understand why we likely cannot embed your product fractal into Euclidean space, let's start in a much simplified setting.  Rather than considering something pathological like a fractal, suppose that $X$ is a smooth manifold.  Don't worry too much about what a smooth manifold is–just keep in mind that a smooth manifold is basically the nicest space you can imagine.  These objects are defined by the fact that if you zoom in enough, they look like Euclidean space.  Examples of smooth manifolds include

*

*an $n$-ball, i.e. $B^n := \{x \in \mathbb{R}^n : |x|<1 \}$,

*an $n$-sphere, i.e. $S^n := \{x \in \mathbb{R}^n : |x|=1\}$,

*an open interval in $\mathbb{R}$,

*the torus, i.e. $S^1 \times [0,1]$, where $S^1$ is the circle, and

*the Klein bottle.

The dimension of a smooth manifold is the dimension of the Euclidean space that you "see" when you zoom in on the manifold.  For example, the two-sphere $S^2$ is a two-dimensional manifold:  if you zoom in enough, small pieces of the sphere look like small pieces of the two-dimensional Euclidean plane.
One of the important results in the study of smooth manifolds is the Whitney embedding theorem.  This theorem asserts that if $X$ is an $m$-dimensional manifold, then $X$ can be embedded into $2m$-dimensional Euclidean space.  Sometimes we can do better than this, but often not.  For example, the circle $S^1$ is a one-dimensional manifold, but it cannot be embedded into the real line (i.e. one-dimensional Euclidean space).  It naturally "lives in" two-dimensional space.  Or consider the Klein bottle, which is a two-dimensional manifold which naturally lives in four-dimensional space–the Klein bottle cannot be embedded into three dimensional space.
The take-away message from this discussion is that even in the nicest possible case, it is often not possible to put a three-dimensional object into three-dimensional Euclidean space, which makes it quite difficult to visualize such objects.  This implies that visualizing a three-dimensional fractal set is likely to be much, much more difficult.
Fractal Sets
With the above in mind, there are some results which might be useful.  Your original fractal set $X$ is naturally a subset of $\mathbb{R}^2$, which means that $X\times X$ naturally lives in $\mathbb{R}^4$.  We do have techniques for visualizing such sets:  for example, we might animate a sequence of cross sections, or consider projections onto three-dimensional hyperplanes.  I honestly don't know what this would do to a Minkowski sausage, but it gives you a place to look.
Another possibility is to consider various embedding theorems.  For example, Mañé shows that if the set
$$ A-A := \{x-y : x,y\in A\} $$
has Hausdorff dimension $n$, then "most" projections of $A$ into $n$-dimensional Euclidean space will be embeddings.  In other words, if you can determine that $A-A$ has Hausdorff dimension less than 3, then almost any projection you choose will embed that set $A$ into $\mathbb{R}^3$, giving you a way of visualizing those sets.
More generally, Hunt and Kaloshin show that similar result holds for sets $A$ with $\dim_f(A-A) < n$ (where $\dim_f$ is the upper box dimension), with the modification that the embeddings are only Hölder continuous (we weaken the hypotheses a little, so the result has a little more wiggle in it).  Olson and Robinson obtain similar results for (almost) bi-Lipschitz maps and the Assouad dimension.
