Remark 4.5.3 in Kalton and Albiac's Topics in Banach Space Theory states:
" ... since $C(K)^*$ is isometric to $\ell_1$ for every countable compact metric space $K$, the Banach space $\ell_1$ is isometric to the dual of many nonisomorphic Banach spaces."
With regard to the above, from H. P. Rosenthal's article in The Handbook of the Geometry of Banach Spaces, Vol 2, a result of Bessaga and Pełczyński is given:
Let $K$ be an infinite countable compact metric space.
$\ \ \ $(a) $C(K)$ is isomorphic to $C(\omega^{\omega^\alpha}+)$ for some countable ordinal $\alpha\ge0$.
$\ \ \ $(b) If $0\le\alpha<\beta<\omega_1$, then $C(\omega^{\omega^\alpha}+)$ is not isomorphic to $C(\omega^{\omega^\beta}+)$
See also this post at MathOverflow which gives an example of a separable Banach space $X$ and a non-separable Banach space $Y$ such that $X^*$ and $Y^*$ are isometrically isomorphic.
Suppose $\mu$ is a (regular Borel) measure such that $L^1(\mu)=C_0(X)^*$, i.e. for every regular Borel measure $\nu$ on $X$ we have a function $f\in L^1(\mu)$ such that for every Borel $A\subseteq X$ we have $\nu(A)=\int_Af(x)\,\mathrm{d}\mu(x)$.
This implies that in particular, this holds for all Dirac measures $\delta_x$ for $x\in X$, which implies that for each $x\in X$ we must have $\mu(\{x\})>0$ and $\mu(\{x\})<\infty$.
Since $\mu(\{x\})<\infty$, by regularity, we conclude that the space $X$ must be locally countable: if every neighbourhood of a point was uncountable, then they would all have infinite measure, and so by regularity, so would $\{x\}$.
I suspect that if $X$ is locally compact and locally countable, then you can find such a measure, though I don't see any obvious construction in general.
Best Answer
Suppose first that $C_0(X) = E^\ast$ for some normed space $E$. By Alaoglu's theorem the closed unit ball $B$ of $C_0(X)$ is compact in the weak*-topology and by the Krein-Milman theorem $B$ has an extreme point (in fact, $B$ is the weak*-closed convex hull of its extreme points).
Thus, in order to prove that $C_0(X)$ is not a dual space, it suffices to show that $B$ has no extreme points:
Let $f \in B$ with $\lVert f\rVert =1$. Since $X$ is not compact and $f$ vanishes at infinity, the set $U = \{x \in X : |f(x)| \lt 1/2\}$ is non-empty open and $C=\{x \in X : \lvert f(x) \rvert \geq 1/2\}$ is non-empty and compact.
Pick $u \in U$ and use Urysohn's lemma to find a function $h\colon X \to [0,1/2]$ such that $h(u) = 1/2$ and $\operatorname{supp}h \subset U$. Then $\left\lVert f \pm h\right\rVert_\infty = 1$ and $f \neq f\pm h$ together with $$ f = \frac{1}{2}\left(f+h\right) + \frac{1}{2}\left(f-h\right) $$ show that $f$ is not an extreme point in $B$.
Remark. It is essential that we assume that $X$ is not compact. Constant functions of norm $1$ are always extremal in the unit ball for compact $X$, so the above argument breaks down. There's a good reason for that: For finite $X$, $C(X) \cong \mathbb{R}^n$ is reflexive, or for $X = \beta\mathbb{N}$ we can show that $C(\beta\mathbb{N}) \cong \ell_\infty = (\ell_1)^\ast$, so these are dual spaces.