Maximum imbalance in a graph?
Since $|e_i|$ is bounded by the diameter $d$, the $\ell_1$ norm is going to be trivially bounded by $nd$, likewise for the $\ell_2$ norm, except by $\sqrt{n}d$ (in fact the $\ell_p$ norm is bounded by $n^{1/p}d$).

The $\ell_1$ case turns out to be surprisingly easy to analyze.

For a path, it's easy to see that $\|\vec e\|_1$ is $O(n^2)$, so you can't do any better than $O(nd)$.

For a complete $k$-ary tree, you can divide it in half at the root, setting $w_{\text{root}} = 0$, ascending one side and descending the other until the leaves have $|e_i| = |w_i| = \log_k n$, producing $O(n\log_k n) = O(nd)$ again.

For a clique it doesn't really matter how you distribute the weights, since they'll all be within $1$ of each other, and that will yield $O(n) = O(nd)$ again.

When you realize that what we're talking about here is a function $e : \mathbb{Z} \to [-d/2,d/2] \subset \mathbb{R}$, and then we're taking its $\ell_1$ norm, as long as you can arbitrarily distribute weights $e_i \in [-d/2,d/2]$ evenly across the range, the bound will be $O(nd)$. 

The only way to change this is to play games with the mass. For instance, if you have several giant cliques at points that are necessarily balanced, like a giant clique with two paths of equal length jutting out of it, then you can count on a bound of only (for example) $O(d^2)$. 

This may be true for expanders to some degree as well, but I'm not sure. I could imagine a case where you set $w_1 = 0$ in a regular graph and then let the values increase subsequently from every hop. It seems likely that the mean could possibly have the most mass, but I don't know if it would be enough to affect the bound.

I think that you could reason similarly about $\ell_2$.

EDIT:

In the comments we figured out a (loose) $\ell_2$ bound of $O(|E|/\lambda_2(L))$ using the constraints of the problem and some basic spectral graph theory.