Edit: My friend helped me to solve this, and now it is solved for me. We must first separate the simple cases, the only serious case is when $\sum_{i=0}^n |f_i|$ and $\alpha$ are both strictly greater than $1$.
Let $L/K$ be an extension of local fields. and let $|.|_L$ be the absolute value on $L$. Suppose we have $K[x] \ni f(x)=\prod_{i=1}^n(x-\alpha_i)$ for $\alpha_i$'s in $L$. I want to show that $|\alpha| < \sum_{i=0}^n |f_i|,$ for every root $\alpha$ of $f$, where $f(x)=\sum_{i=0}^n f_ix^i$. It is about an hour that I stuck in it without no progress, what's make it worser is that I don't have even a raw idea.
Let $m := \max \lvert \alpha_i \rvert$, and let $ r \in \{1, .., n\}$ be the number of $\alpha_i$'s such that $\lvert \alpha_i \rvert = m$. Then note that by the ultrametric maximum principle, $\lvert f_{n-r} \rvert = m^r$. Hence
$$ \sum \lvert f_i \rvert \ge \lvert f_n \rvert + \lvert f_{n-r} \rvert = 1+m^r > m $$
(the last inequality is true for any real $m \ge 0$), which by definition is $\ge \lvert \alpha_i \rvert$ for all $i$.