True. For any $n\in \mathbb{N}$  consider the inclusion to the $n$-th coordinate   $j _ n : X\to c _ 0(X)$ which is right inverse to the evaluation at $n$, so that $(j _ n x)(n)= x$, for any $x\in X$. Let  $j _ n ^ T : c _ 0(X) ^ * \to X^*$ its transpose operator. Any $\eta \in c _ 0(X)^ * $ defines a sequence $y:\mathbb{N}\to X ^ *$ such that $y(n)   := j _ n ^T \eta $. 
The $\ell _ 1(X^*)$- norm of $y$ is 
$$\|y\|_ { \ell _ 1 (X^*)}=\sum _{n\in\mathbb{N}}\\  \|y(n)\| _ {X ^ *} = \sum _{n\in\mathbb{N}}\\ \\ \sup _ {\|x\| _ X \le 1} \langle y(n), x  \rangle=$$$$ = \sup _ {m\in\mathbb{N}}\\ \\ \sup _ {\|\xi\| _ { c _ {0} (X)} \le 1} \\ \sum _{n\le m}\\ \langle y(n),   \xi(n)  \rangle =$$$$=  \sup _ {\|\xi\| _ { c _ {0} (X)} \le 1} \\ \langle \eta,  \xi  \rangle = \|\eta\| _  {c _ 0(X)^*}\\ \\ .$$
This shows that the inclusion $\ell _ 1(X^*)\to c _ 0(X)^ * $ is actually a linear (surjective) isometry.