Let $G$ be a finite group, for each irreducible character $\chi$, we define ${\bf Z}(\chi)$ to be the set of all $x\in G$ such that $|\chi(x)|=\chi(e)$ when $e$ is the identity of the group.
For every irreducible charcter we know that 


$\frac{|G|}{\chi(e)^2} \leq \frac{1}{\chi(e)} \sum\_{x\in G} |\chi(x)| \leq \frac{|G|}{\chi(e)}$.


Now suppose that ${\bf Z}(\chi)$ is trivial i.e. it includes only $e$. Can we find a better bound for
$\frac{1}{\chi(e)} \sum\_{x\in G} |\chi(x)|$?


My firs clue about this question was in observance of affine $p$-groups: they satisfy this condition: (${\bf Z}(\chi)$ is trivial for only non-linear character $\chi\_\pi$ of affine $p$-groups). On the other hand, for this non-linear character, 


$\frac{1}{\chi\_\pi(e)} \sum\_{x\in G} |\chi\_\pi(x)| =2$


for every $p$. 


Subsequently, I tried some more character tables for finite groups that some of their characters, say $\chi$, satisify this condition and I observed that $\frac{1}{\chi(e)} \sum\_{x\in G} |\chi(x)|$ is so close to $\frac{|G|}{\chi(e)^2}$ than $\frac{|G|}{\chi(e)}$.



Some special characters of finite groups
The rightmost inequality can be improved for all non-linear irreducible characters, and if $Z(\chi) = 1$ and $G$ is non-trivial, then $\chi$ must certainly be non-linear. Using Cauchy-Schwarz, we have $$\left(\sum\_{x \in G} |\chi(x)|\right)^{2} \leq m(\chi) \left(\sum\_{x \in G} |\chi(x)|^{2}\right),$$ where $m(\chi)$ is the number of group elements where $\chi$ does not vanish (so certainly $m(\chi) \leq |G|$, and the inequality is strict when $G$ is non-linear,
since by a Theorem of Burnside, each non-linear irreducible character vanishes somewhere). Hence $\sum\_{x \in G}|\chi(x)| \leq \sqrt{m(\chi)|G|}$, which removes (at least) a factor of $\chi(1)$ from your inequality.


When $Z(\chi) = 1$, the leftmost inequality becomes strict if $G$ is non -trivial.
Added later: it might be worth pointing out that Burnside's argument also implictly shows
that $\sum\_{x \in G}|\chi(x)| > m(\chi)$ when $\chi$ is a non-linear irreducible character (just notice that if $\chi(x) \neq 0$, then $1 \leq \prod\_{y} |\chi(y)|$, where $y$ runs through all the generators of $\langle x \rangle$, and use the arithmetic-geometric mean
inequality). 


