OP's question was about being *isomorphic* to a dual space so we need to observe that $L_1$ lacks the Radon–Nikodym property, which is invariant under isomorphisms, and separable dual spaces have this property.

Also, the $\ell_1$-sum of continuum many copies of $L_1[0,1]$ is isometric to $C[0,1]^*$.

*Edit of 31.07.2016*: It has been pointed out that my answer is incomplete as I do not treat the case when $L_1(\mu)$ is non-separable for a $\sigma$-finite measure $\mu$. By the Radon–Nikodym theorem, we may assume without loss of generality that $\mu$ is actually *finite*.

The argument is then almost exactly the same as in this case the inclusion $L_2(\mu)\subset L_1(\mu)$ has dense range so $L_1(\mu)$ is weakly compactly generated. It is well known that weakly compactly generated dual spaces have the Radon–Nikodym property, a property that $L_1(\mu)$ is clearly lacking (by Maharam's theorem, $L_1(\mu)$ is isometric to $L_1(\{0,1\}^\lambda)$, where $\{0,1\}^\lambda$ is considered with the product fair-coin-toss (Haar) measure and $\lambda$ is the density character of $L_1(\mu)$).