What is the predual of $L^1$

In fact, $L_1[0,1]$ has no pre-dual. More is true: $L_1$ cannot be embedded is a separable dual space. See, e.g., Theorem 6.3.7 in Kalton and Albiac's Topics in in Banach Space Theory.

Banach-Alaoglu-Bourbaki: For $X$ a Banach space, the closed unit ball of $X^*$ is weak*-compact.

Krein-Milman: For $X$ a locally convex topological vector space, and $K$ a compact convex subset, $K$ is the closed convex hull of its extreme points.

Suppose $X^* = L^1[0,1]$. By BAB, the closed unit ball of $L^1[0,1]$ is weak*-compact. In addition, weak* topologies are always locally convex (see, for instance, here) and $L^1[0,1]$ is clearly convex so Krein-Milman gives that it is the closed convex hull of its extreme points.

But the closed unit ball of $L^1[0,1]$ has no extreme points: Suppose $\|f\|_1 = \alpha \leq 1$, $f \neq 0$. Then $H(s) = \int_0^s |f(t)| \, dt$ is a continuous function, with $H(0) = 0$ and $H(1) = \alpha$. Thus there is some $s_0 \in (0,1)$ with $H(s) = \frac{\alpha}{2}$, so set $$g(t) = 2f \chi_{[0,s_0]} \quad h(t) = 2f \chi_{[s_0,1]}$$ which satisfy $\|g\|_1 = \|h\|_1 = \|f\|_1 = \alpha$ and $\frac{1}{2}(g + h) = f$, so $f$ is not an extreme point. Clearly $0$ is not an extreme point either, so the unit ball of $L^1[0,1]$ has no extreme points.

Thus we have a contradiction, and $L^1[0,1]$ has no pre-dual.