I understand that $$ \|f\|_{E^*}=sup_{\|x\|_E=1}|f(x)|. $$ I am confused about $f_i=\langle f,e_i\rangle$. Since $f\in E^*$ it is a functional $f:E\rightarrow \mathbb{R}$. My best guess is since $x=x_i e_i$ we can think of $f$ as $f(x)=f(x_1,x_2,...x_n)$.
The answer is $$\|f\|=\max_i|f_i|$$ Proof. Given a functional $f\in E^*$ and $x\in E$, write $x=\sum_{I=1}^nx_ie_i$, so that $$f(x)=f\left(\sum_{i=1}^nx_ie_i\right)=\sum_{I=1}^nx_if_i$$ so $$\|f\|=\max\{|f(x)|: \|x\|_1=1\}=\max\{|\sum_{i=1}^nx_if_i|:\sum_{i=1}^n|x_i|=1\}=\max_i|f_i|$$ To prove this equality, note that for every choice of $x_i$ such that $\sum_{i=1}^n|x_i|=1$, we have $$|\sum_{i=1}^nx_if_i|\leq \sum_{i=1}^n|x_if_i|\leq \sum_{i=1}^n|x_i|\cdot\max|f_i|\leq\max_i|f_i|$$ Also, given $f\neq 0$, let $j$ be an index such that $|f_j|=\max_i|f_i|$. Since $f\neq 0$, we have $f_j\neq 0$. Define a vector $x\in E$ by $x_j=|f_j|/f_j$ and $x_i=0$ for $i\neq j$. Then $\|x\|_1=1$ and $f(x)=|f_j|=\max_i|f_i|$, so that there is equality as claimed.