Why not define $|v| = -1$? I was wondering why if we have $i^2 = -1$, why not have a "number" $v$ such that $|v| = -1$?  Does anything interesting arise from considering this system?
The only thing I could come up with was:
$$ |a+bv| \le |a| + |bv| = a - b $$
 A: Such a number (or vector) $v$ simply can't exist if you want $|{-}|$ to behave like $|{-}|$ should behave. For example we'd have
$$|1+v| \le |1|+|v| = 1-1 = 0$$
and so
$$1 = |1| = |(1+v)-v| \le |1+v|+|(-v)| = |1+v|+|v| \le 0 + (-1) = -1$$
But $1 \le -1$ is nonsense.
In general there is no reason why arbitrarily postulating the existence of a 'thing' satisfying a certain property should yield anything worthwhile. The story behind $i$ is somewhat different.
A: First off, it wouldn't be very interesting to have a norm that was negative for all nonzero vectors in a vector space -- that would just be the same old norm we already know, with a minus in front of it.
So suppose we have vectors $v, w$ with $|v|<0$ and $|w|>0$. Let's further assume that $v$ and $w$ are linearly independent. If they are not, assuming that our vector space is at least two dimensional, let $q$ be some vector that is linearly independent of $v$ (and this also $w$). If $|q|>0$ then use $q$ as the new $w$; otherwise use $q$ as the new $v$.
Now consider the function
$$ f(t) = |(1-t)v + tw| $$
This is certainly something that looks like it ought to be continuous on $[0,1]$, but $f(0)=|v|<0$ and $f(1)=|w|>0$, so by the intermediate value theorem, there should be some $t$ such that $|(1-t)v+tw|=0$. And because $v$ and $w$ are linearly independent, $(1-t)v+tw\ne 0$.
Now we have a nozero vector with zero norm, and that wreaks all kinds of havoc for most of the things we want to use norms for.

Bonus chatter: There are in fact spaces with norm-like things $\|{\cdot}\|$ such that $\|v\|^2$ is always real, but can be either positive or negative(!). A prominent example is Minkowski space, which forms the "spacetime" of Special Relativity. Here we do indeed have nonzero vectors with zero $\|v\|$, which play a special role in the theory -- but for that very reason they don't satisfy many of the properties the usual length of a vector has.
A: The absolute value of a real or complex number is necessarily a non-negative real number (this is a property of norms), and so breaking this property means most of the usual statements involving $|\cdot|$ and $v$ will probably be false. 
If you wanted to, however, there's no reason you couldn't take some abstract set $E$, containing (at least) your new "number" $v$, and investigate some function 
$$\phi : E \cup \Bbb C \to \Bbb R$$
such that $\phi(z) = |z|$ for $z \in \Bbb C$ and $\phi(v) = -1$ for some $v \in E$ (and then you have to decide what else $\phi$ does or what properties you want it to have). 
