Given a Hilbert space.
Symmetric operators can be described by $$\overline{\mathcal{D}(A)}=\mathcal{H}:\quad A\subseteq A^*\iff\langle A\varphi,\psi\rangle=\langle\varphi,A\psi\rangle\quad(\varphi,\psi\in\mathcal{D}(A))$$
So that essentially selfadjoint operators can be characterized by: $$A\subseteq A^*:\quad\overline{A}=A^*\iff A^*\subseteq A^{**}$$ (The existence of adjoints being implicitely required.)
Does it fail if symmetry assumption on the original operator is dropped?
More short, can it happen: $A^*\subsetneq A^{**}=\overline{A}$
Let $B : \mathcal{D}(B)\subset X \rightarrow X$ be any closed densely-defined linear operator on a Hilbert space $X$, with $B \prec B^{\star}$ (I am using $\prec$ to denote strict graph inclusion.) Then $A=B^{\star}$ satisfies $A^{\star} \prec A^{\star\star}$. Indeed, because $B$ is closed and densely-defined, then $B^{\star\star}=\overline{B}=B$, which gives the strict graph inclusion $$ A^{\star} =(B^{\star})^{\star}=B \prec B^{\star}= A=\overline{A}=A^{\star\star}, $$ and is as you stated: $$ A^{\star} \prec A^{\star\star} = \overline{A}. $$ An example is $X=L^{2}[0,\infty)$ and $B=-i\frac{d}{dx}$ defined on the domain $\mathcal{D}(B)$ consisting of all absolutely continuous functions $f \in X$ for which $f' \in X$ and $f(0)=0$. Then, $$ (Bf,g) = (f,Bg),\;\;\; f,g\in \mathcal{D}(B). $$ The operator $B$ is closed and densely-defined with adjoint $B^{\star}=-i\frac{d}{dx}$ on the domain $\mathcal{D}(B^{\star})$ consisting of all absolutely continuous $f \in X$ for which $f' \in X$ (without any endpoint condition at $0$.) The opertor $B$ is symmetric, but $A=B^{\star}$ is not.