Is the intuition that the fractal dimension measures the extent of space filling not valid for box fractals? 
Box fractal refers to various iterated fractals created by a square or
rectangular grid with various boxes removed or absent and, at each
iteration, those present and/or those absent have the previous image
scaled down and drawn within them. from Wikipedia

Consider three different box fractals.

*

*One square is removed from a 2x2 grid, so that 25% of the space is removed. The fractal dimension is $\frac{\log(3)}{\log(2)}\approx 1.585$. This is equivalent to the Sierpinski triangle.

*Two squares are removed from a 2x3 grid, so that 33% of the space is removed. This fractal shape is referred to as a "McMullen carpet" (refs: 1, 2). The fractal dimension is $\frac{\log(2)}{\log(2)} + \frac{\log(4/2)}{\log(3)} \approx 1.631$. This is equivalent to removing three squares from a 3x3 grid.

*Four squares are removed from a 3x3 grid, so that 44% of the space is removed. This is the Vicsek fractal, with a dimension of $\frac{\log(5)}{\log(3)}\approx 1.465$
Even though fractal #2 removes more space than fractal #1, its box-counting dimension is higher. However, higher fractal dimensions are usually associated with fractals which fill more space. Is there an intuitive way to understand why fractal #2 has a higher fractal dimension despite filling less space?
 A: I think there's really just more to the concept of fractal dimension in the context of self-affine sets than there is in the context of self-similar sets. In particular, the dimension is much more sensitive to the placement of the pieces and not just "how much is removed".
Consider the self-affine set implied by the following three rectangular images of the unit square:

If we iterate this a few times, we get an image that looks like so:

It seems reasonably clear that the fractal dimension of this object is larger than 1. In fact a formula of McMullen described in this answer indicates that the dimension is
$$
\frac{\log(2)}{\log(2)} + \frac{\log(3/2)}{\log(3)} \approx 1.36907.
$$

Now consider the following variation implied by this collection of rectangles:

We are again removing three of the six rectangles from a $2\times3$ decomposition of the unit square so we've certainly removed the exact same amount and will have the same area at every step. Now, the same formula implies that the dimension is
$$
\frac{\log(1)}{\log(2)} + \frac{\log(3/1)}{\log(3)} = 1.
$$
More importantly, in terms of the intuition that you seek, it is easy to see why the dimension is $1$. Just take a look at the next step:

The width of the entire set is contracted by the factor $1/2$ with each step. As a result, the attractor is really just a horizontal line segment, which (of course) has dimension 1:


Personally, I think it's still reasonable to think that a fractal with larger dimension is better at "filling up space" than one with a smaller dimension. Another key issue is that "degree to which space is filled up" is not really preserved under the limiting operation. That is, two sequences of sets with the same area might approach very different limiting objects.
