No
Your idea is good, but you are misunderstanding what 'any dimension' means.
When you think of a 1-foot cube, you are thinking that, since the cube is no longer than 1 foot to a side, it has a length of 1 foot in each direction. This is not true. In fact, the cube has an extent of no less than one foot in each direction and no more than √3.
In order for total extent of the cloak to be no longer than 1 foot in any dimension, you actually need, at best (in terms of volume), a 1-foot diameter sphere. If you are using a cube it would need to be a 1/√3-foot cube.
That said, even this is not enough.
Imagine your tear is a simple 2-foot line. Let's say you fold it in half, so now you have a line that goes 1 foot in a direction and then another 1 foot back to where it started. Being able to rely on the volumetric simplification described above means being able to rely on the folded tear being considered to be 1-foot in length. This is by no means the most likely interpretation of 'length'-- that would be that the total distance traveled along that axis is 2-feet, so the length of the path in that dimension is, similarly, 2 feet. Otherwise the intuitive theorem
$$\sum_{i=0}^n length_i = total \; length$$
is, in fact, false.
If the length of a line folded exactly upon itself does not change, then there is a maximum to the length of a single-line tear that can be mended since folding can only accomplish so much and travel in non-cardinal directions is not unrelated. Travel in any direction counts as travel in any other direction with the length reduced by multiplying by the cosine of the angle between them, which makes the math complicated.
You can mend a tear of pi/2 feet in length by folding it into a perfect circle, and this is the theoretical maximum in two dimensions. You can add more on for each extra dimension.
But that's too hard/long for playing!
I mean, that depends on the group. But if it doesn't work, just remember wizards can use origami+complicated math to increase by a substantial but less-than-doubling amount how much they can mend, and if they can go to planes with more spatial dimensions then they can fix more with even more complicated folds and accompanying geometric analyses. The exact amount doesn't need to be calculated any more than how big the hole in the wizard's cloak needed to be when you put it there.
Proof:
The theoretical maximum is achieved when the total length in each dimension exactly equals 1 foot, if such a curve exists.
An interesting property of circles is that they are perfectly identical regardless of how you rotate them; a circle is the same shape with the same relevant properties regardless of what orientation you pick to use as an axis. Because of this property, the 'length' of a circle in each and every dimension is the same.
Another helpful property is that, for any given dimension, we can break a circle up into two always-increasing or always-decreasing half-circles. This lets us sum the length for each and then combine them into a total length in that dimension, rather than having to take a line integral (it's an easy line integral though, and good practice if you are learning such things).
For always-increasing (or decreasing) curves, the 'length' as we have defined it is identical to the magnitude of the displacement because:
$$ \Delta x = \sum_{i=0}^n dx_i$$ where dx is the signed infinitessimal displacement vector at x=i, and the length is $$ s_x=\sum_{i=0}^n |dx_i| $$, but for always-positive dx in the always-increasing case we can remove the absolute value leaving $$s_x=\sum_{i=0}^n dx_i \cong \Delta x$$
So, the 'length' of a circle in a single dimension is twice its diameter. Since we can't exceed 1 foot, that gives $\pi d= \pi /2$ for the perimiter of the circle. We've know we've used up all our possible length in two dimenstions, so this result must be optimal.
"Wait!" You might think. "What if we used many circles? Each circle still uses the maximum amount of 'length' in each direction, so we should be able to combine them to make a thing, too!"
Yep! That's accurate, but the thing isn't any bigger-- it's just equally optimal. You can fold your tear into one big circle or a very circular infinity sign or the olympic rings or whatever you want, as long as it's all circles, and you'll still get exactly pi/2 length at best. Yay math!
You do get extra length in three dimensions (and additional length for each dimension you add after that), but the math is way more complicated so I need to work on making a good, simple explanation for that. Feel free to edit one in if so inclined!
