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.