I was wondering about the same question and was surprised that I could not find an answer to this anywhere. The answer to your question is yes.

Let $(X, \|\cdot\|)$ be a normed vector space and $Y \subset X$ a closed subspace. Then $(X / Y)^*$ is canonically isomorphic to $Y^\perp$, where $Y^\perp = \{\varphi \in X^* : \ker(\varphi) \subset Y \}$ is the annihilator of $Y$. The isomorphism is given by
\begin{align}
&\Phi : Y^\perp \rightarrow (X / Y)^*, \\
&\Phi(\varphi)([x]) = \varphi(x).
\end{align}

Hence, if a subset $F \subset X^*$ satisfies $F = Y^\perp$ for some $Y$, then $$ (X/ Y)^* \cong F, $$ i.e. $F$ has a predual. Given $F$, a natural candidate for $Y$ is $F^\perp = \{x \in X : x \in \ker(\varphi) \text{ for all } \varphi \in F \}$, the pre-annihilator of $F$. However, in general it is not true that $$ F = (F^\perp)^\perp \tag{1}\label{1}. $$ Fortunately, $W^{1, \infty}(\Omega)$ satisfies this property for a suitable choice of $X$. For the moment, assume $(1)$ holds. Given a sequence $(\varphi_k) \subset F$, we compare weak$^*$ convergence in $X^*$ and $F$. Since $(X / F^\perp)^* \cong F$, $\varphi_k \overset{\ast}\rightharpoonup \varphi$ in $F$ means
\begin{align}
\Phi(\varphi_k) \overset{\ast}\rightharpoonup \Phi(\varphi) \text{ in } (X / F^\perp)^* \iff &\Phi(\varphi_k)([x]) \to \Phi(\varphi)([x]) \quad \forall [x] \in X / F^\perp \\
\iff &\varphi_k(x) \to \varphi(x) \quad \forall x \in X.
\end{align}
But this is the definition of weak$^*$ convergence in $X^*$, so we are done.

It remains to show that $W^{1, \infty}(\Omega)$ satisfies $(1)$. Let $X = L^1(\Omega, \mathbb{R}^{n + 1})$ and $F = W^{1, \infty}(\Omega)$. We think of $F$ as a subset of $X^* = L^\infty(\Omega; \mathbb{R}^{n + 1})$ by identifying $u \in W^{1, \infty}(\Omega)$ with $(u, D_1u, \dots, D_nu) \in L^\infty(\Omega, \mathbb{R}^{n + 1})$. By definition, $F^\perp$ consists of all $(f_0, \dots, f_n) \in L^1(\Omega, \mathbb{R}^{n + 1})$ such that $$ \int_\Omega u f_0 + \sum_{j = 1}^n D_ju f_j \, dx = 0 \quad \forall u \in W^{1, \infty}(\Omega).$$ Suppose $v \in (F^\perp)^\perp \subset L^\infty(\Omega, \mathbb{R}^{n + 1})$, i.e. $$ \int_\Omega \sum_{j = 0}^n v_j f_j \, dx = 0 \quad \forall f \in F^\perp.$$ We prove that $v \in F$ by showing that it is weakly differentiable. For $\eta \in C_c^\infty(\Omega; \mathbb{R}^n)$ we define a function $f \in L^1(\Omega; \mathbb{R}^{n + 1})$ by
\begin{align}
f_0 &= \text{div}\, \eta, \\
f_j &= -\eta_j \quad j = 1, \dots, n.
\end{align}
Observe that $f \in F^\perp$. This follows simply from the definition of the weak derivative. Hence, $$ 0 = \int_\Omega \sum_{j = 0}^n v_j f_j \, dx = \sum_{j = 1}^n \int_\Omega v_0 D_j\eta_j + v_j \eta_j \, dx \implies v_j = D_jv_0 \quad j = 1, \dots, n.$$

double stars`**double stars**`

rather than $\mathbf{\text{math mode}}$`$\mathbf{\text{math mode}}$`

. I have edited accordingly. $\endgroup$