The answer to this depends on the field you're in. If you're a mathematician, then [all norms in finite dimensions are equivalent][1]: for any two norms $\|\cdot\|_a$ and $\|\cdot\|_b$, there exist constants $C_1,C_2$, which depend only on dimension  (and a,b) such that:

$$C_1\|x\|_a\leq \|x\|_a\leq C_2\|x\|_b.$$

This implies that norms in finite dimensions are quite boring and there is essentially no difference between them except in how they scale. [This usually means that you can choose the most convenient norm for the problem you're trying to solve][2]. For example, in numerical linear algebra, the Frobenius norm is sometimes preferred because it's a lot easier to calculate than the euclidean norm, and also that it naturally connects with a wider class of [Hilbert Schmidt operators][3]. Also, like the Euclidean norm, it's submultiplictive: $\|AB\|_F\leq \|A\|_F\|B\|_F$, unlike say, the max norm, so it allows you to easily talk about operator multiplication in whatever space you're working in. . 

For *practical* purposes, the differences between norms become more pronounced because we live in a world of dimensions and it usually matters how big a certain quantity is, and how it's measured. Those constants $C_1,C_2$ above are not exactly tight, so it becomes important just how much more or less a certain norm $\|x\|_a$ is compared to $\|x\|_b$. 


  [1]: http://www.win.tue.nl/~rvhassel/Onderwijs/Old-Onderwijs/2DE08-0910/ConTeXt-OWN-FA-120909-Bib/Literature/Equivalent-Norms-findim/norm_equivalence_much-questions.pdf
  [2]: https://en.wikipedia.org/wiki/Matrix_norm
  [3]: https://en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator