I'm reasking this deleted question because I believe I've made a some progress towards an answer, which I'm also interested in knowing.
Here's the restatement:
Suppose $\ A\ $ is a densely defined, [unbounded]$\,^{\color{red}\dagger}$, symmetric, positive operator on a Hilbert space. Is there any additional condition in terms of $\ A\ $ that forces the adjoint $\ A^*\ $ to be also positive?
$\,^{\color{red}\dagger}$ This qualification was omitted from the original statement of the question, although it did appear in its title. If $\ A\ $ is bounded, then it's easy to show that $\ A^*\ $ is an everywhere defined, bounded and positive extension of $\ A\ $, so I'm only interested in the question for $\ A\ $ unbounded.
I first need to deal with one of the comments on the original question. It is not true in general that $\ A^*=A\ $. What is true is that $\ A^*\supseteq A\ $, but it's quite possible that $\ A^*\supsetneq A\ $—i.e. that the extension is strict. The identity $\ A^*=A\ $ holds if and only if $\ A\ $ is self-adjoint.
Update:
I have now found that a necessary and sufficient condition for $\ A^*\ $ to be positive is that $\ A\ $ be essentially self-adjoint and $\ A^*\ $ be its unique self-adjoint extension. A proof is given in the answer below. Although there could well be more of interest that might be added, this necessary and sufficient condition is enough to satisfy my curiosity about this question.
Best Answer
A necessary and sufficient condition for $\ A^*\ $ to be positive is that $\ A^*=A^{**}\ $—that is, the adjoint of $\ A\ $ is self-adjoint. In fact, $\ A\ $ must be essentially self-adjoint, and $\ A^*\ $ must be the unique self-adjoint extension of $\ A\ $.
If $\ A^*\ $ is self-adjoint, then its spectrum is a subset of the real line, and the operators $\ A^*+iI\ $ and $\ A^*-iI\ $ therefore have bounded inverses.
It follows that $\ \ker\big(A^*+iI\big)= \ker\big(A^*-iI\big)=\{0\}\ $, and hence that \begin{align} \mathscr{R}(A+iI)^\perp&=\ker(A^*-iI)=\{0\}\\ \mathscr{R}(A-iI)^\perp&=\ker(A^*+iI)=\{0\}\ . \end{align} Thus, $\ A\ $ is essentially self-adjoint because both its deficiency indices are zero. Since $\ A\ $ therefore has a unique self-adjoint extension, its adjoint, its Friedrichs extension and its Krein-von Neumann extension must all coincide. The adjoint must therefore be positive.
On the other hand, if $\ A^*\ $ is positive, then it is symmetric, and so $\ A^{**}\supseteq A^*\ $. Also, if $\ x\in\mathscr{R}(A+iI)^\perp=$$\,\ker(A^*-iI)\ $, then $$ \langle A^*x,x\rangle=i\|x\|^2\ , $$ which implies that $\ x=0\ $ because $\ A^*\ $ is positive. Therefore, $$ \mathscr{R}(A+iI)^\perp=\{0\}\ , $$ and the deficiency indices of $\ A\ $ (which must be equal because $\ A\ $ is positive symmetric) are zero. So again $\ A\ $ is essentially self-adjoint. Let $\ B\ $ be the unique self-adjoint extension of $\ A\ $. Since $\ B\supseteq A\ $, then $\ B=B^*\subseteq A^*\ $, and therefore, since $\ B\ $ is symmetric, $\ B^*\supseteq A^{**}\ $. We now have the chain of extensions $$ \ A^{**}\supseteq A^*\supseteq B\supseteq A^{**}\ , $$ which imply that $$ A^{**}=A^*=B $$ and hence that $\ A^*\ $ is self-adjoint.
That essential self-adjointness isn't sufficient by itself is shown by the following example of an essentially self-adjoint positive operator, defined on a dense subspace of $\ \mathscr{L}^2[0,1]\ $, whose adjoint is not positive.
Define $$ (Af)(x)=-f''(x) $$ for all twice-differentiable functions $\ f:[0,1]\rightarrow\mathbb{C}\ $ satisfying $\ f'(0)=f'(1)=f(0)=f(1)=0\ $ and $\ \int_0^1|f''(x)|^2\,dx<$$\,\infty\ $. This space of functions is dense (with respect to the $\ \mathscr{L}^2\ $ norm) in $\ \mathscr{L}^2[0,1]\ $. Integration by parts gives \begin{align} \langle Af,f\rangle&=-\int_0^1f''(x)\overline{f(x)}\,dx\\ &=\int_0^1|f'(x)|^2\,dx\\ &\ge0 \end{align} for all $\ f\in\mathscr{D}(A)\ $, so $\ A\ $ is positive. Now if $\ g:[0,1]\rightarrow\mathbb{C}\ $ is defined to be $$ g(x)=x^2\ , $$ then $\ g\in\mathscr{L}^2[0,1]\ $ and \begin{align} \langle Af,g\rangle&=-\int_0^1f''(x)x^2\,dx\\ &=2\int_0^1f'(x)x\,dx\\ &=-2\int_0^1f(x)\,dx\\ &=\langle f,h\rangle \end{align} for all $\ f\in\mathscr{D}(A)\ $, where $\ h(x)=-2\ $ for $\ x\in[0,1]\ $. It follows that $\ g\in\mathscr{D}\big(A^*\big)\ $ and $\ A^*g=h\ $. Now \begin{align} \langle A^*g,g\rangle&=\int_0^1-2x^2\,dx\\ &=-\frac{2}{3}\ , \end{align} and therefore $\ A^*\ $ is not positive.
The essential self-adjointness of $\ A\ $ follows from the following properties: \begin{align} \mathscr{R}(A+iI)^\perp&=\{0\}\\ \mathscr{R}(A-iI)^\perp&=\{0\}\ , \end{align} which are not difficult to prove, and imply that both deficiency indices of $\ A\ $ are $\ 0\ $.