Consider the following operator
$$F:\mathbb{C}^n\to V$$
$$F(a_1,\ldots,a_n)=a_1\cdot v_1+\cdots +a_n\cdot v_n$$
We consider $\mathbb{C}^n$ with taxi cab norm. Now let
$$G:\mathbb{C}^n\to\mathbb{R}$$
$$G(v)=\lVert F(v)\lVert$$
Obviously $G$ is continous (because $F$ is as a finite dimensional operator). Let $\mathbb{S}^n=\{v\in\mathbb{C}^n\ |\ \lVert v\lVert=1\}$ be an $n$-dimensional sphere. Obviously $G$ achieves minimum on $\mathbb{S}^n$ (because $\mathbb{S}^n$ it is compact). Denote this minimum by $c$. Then you have
$$\lVert F(v)\lVert=\lvert v\lvert \cdot \bigg\lVert F\bigg(\frac{v}{\lvert v\lvert}\bigg) \bigg\lVert \geq \lvert v\lvert\cdot c$$
Let $v=(a_1,\ldots, a_n)$. Since we have taxi cab norm in $\mathbb{C}^n$ then
$$\lVert F(a_1,\ldots, a_n)\lVert\geq c\cdot\sum_{i=1}^n\lvert a_i\lvert$$
This is the inequality you've asked us to look at.
So in order to find $c$ you just have to find infimum of $G$ on $\mathbb{S}^n$. In other words
$$c=\inf\bigg\{\bigg\lVert \sum a_i\cdot v_i\bigg\lVert\mbox{, where}\sum\lvert a_i\lvert=1\bigg\}$$
So this is strict equality and there is no better choice of $c$.
Now by taking various points from the sphere (remember that there is the taxi cab norm on $\mathbb{C}^n$) you can find some interesting inequalities. For example for $(a_1,\ldots,a_n)=(0,\ldots,1,\ldots, 0)$ where $1$ is on $i$-th place and zeros otherwise you get
$$c\leq \lVert v_i\lVert$$
for all $i$.