A natural-looking distance formula The distance formula in one dimension is
$$D_1 = |x_2-x_1|$$
From the Pythagorean theorem, the distance formula in two dimensions is
$$D_2 = \sqrt{|x_2-x_1|^2 + |y_2-y_1|^2}$$
Now, in three dimensions, a student might innocently suppose that
$$D_3 = \sqrt[3]{|x_2-x_1|^3 + |y_2-y_1|^3 + |z_2-z_1|^3}$$
yet the Pythagorean theorem shows otherwise.  Is there any physical intuition behind distances defined by $D_3$?  What strange type of ruler would be needed to measure such a thing?
 A: Instead of talking about distances $D(x,y)$ it is slightly easier to talk about norms $$N(x-y).$$
Your proposes $D_3$ is called the $L_3$ norm and is part of a family of norms
$$\|x\|_p = \left(|x_1|^p + \ldots + |x_n|^p\right)^{1/p}$$
called the $L_p$ norms. For $p\geq1$ they obey all of the properties needed for a norm: positive for nonzero $x$, linear under multiplication by positive scalars, and satisfy the triangle inequality.
$p=2$ is the usual Euclidean norm and induces the usual Euclidean distance. In one dimension, all of the $L_p$ norms are equivalent, which is why you interpreted the $L_2$ norm as the $L_1$ norm.
So to answer your question, nothing goes wrong when you use the $L_3$ norm. You will get lengths that are somewhere between using the usual $L_2$ norm, and the $L_{\infty}$ or "max" norm which returns the magnitude of the largest component; as such, components of a vector that are small relative to the largest components will tend to get ignored by an $L_3$ ruler.
Here's what the unit circle would look like using such a norm:

