Is the $L^1$-space dual to a Banach space

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 $L_1(\mu)$).

I would like to mention that for general measure spaces the answer is negative: The space $L_1(0,1)^{**}$ (by, e. g., Proposition II.5.3 in Lindenstrauss--Tzafriri (1973)) is a dual $L_1$-space.

Actually for purely atomic (even finite) measures the negative answer is obvious: consider a probability measure on a countable set.