That's correct. Your assumptions are $(T_nx,y)\longrightarrow (Tx,y)$ for all $x,y$ (sot hence wot limit). And $(T_nx,y)=(x,T_ny)$ for all $x,y$ and all $n$ (self-adjoint). So
$$(Tx,y)=\lim (T_nx,y)=\lim (x,T_ny)= (x,Ty)$$
for all $x,y$. That is: $T$ is self-adjoint. The first and the third equalities are from the wot condition. The middle equality is by self-adjointness of each $T_n$.
The sequence $(T_{n}x,x)$ is convergent: $(T'x,x)\geq(T_{n+1}x,x)\geq(T_{n}x,x)$ for each $n=1,2,...$
Then sequence $(T_{n}x,y)$ is also convergent by means of polarization.
We can define $(Tx,y)=\lim_{n}(T_{n}x,y)$. It is easy to see that $T$ is self-adjoint because of those $T_{n}$.
On the other hand, by Uniform Boundedness Principle, one can show that $\|T_{n}\|\leq C$ and hence $T$ is also bounded.
Indeed, consider the sesquilinear form $U(x,y)=(T_{n}x,y)$, then we have
\begin{align*}
\|T_{n}x\|^{2}&=U(x,T_{n}x)\\
&\leq U(x,x)^{1/2}U(T_{n}x,T_{n}x)^{1/2}\\
&=(T_{n}x,x)^{1/2}(T_{n}^{2}x,T_{n}x)^{1/2}\\
&\leq(T_{n}x,x)^{1/2}(T'(T_{n}x),T_{n}x)^{1/2}\\
&\leq(T_{n}x,x)^{1/2}\|T'\|^{1/2}\|T_{n}x\|,
\end{align*}
so $\|T_{n}x\|\leq(T_{n}x,x)^{1/2}\|T'\|^{1/2}\leq(T'x,x)^{1/2}\|T'\|^{1/2}$, so for each $x$, $\|T_{n}x\|$ is bounded, now we apply Uniform Boundedness Principle.
Now consider the sesquilinear form $S(x,y)=((T-T_{n})(x),y)$.
We have
\begin{align*}
\|(T-T_{n})x\|^{2}&=S(x,(T-T_{n})x)\\
&\leq S(x,x)^{1/2}S((T-T_{n})x,(T-T_{n})(x))^{1/2}\\
&=((T-T_{n})(x),x)^{1/2}((T-T_{n}x)^{2},(T-T_{n})(x))\\
&\leq((T-T_{n})(x),x)^{1/2}\|(T-T_{n})(x)\|^{1/4}\|(T-T_{n})^{2}(x)\|^{1/4}\\
&\leq C'((T-T_{n})(x),x)^{1/2}\|x\|^{1/2}\\
&\rightarrow 0.
\end{align*}
Edit:
The boundedness of $\|T_{n}\|$ can be proved without appealing to $T'$.
Indeed, the sequence $(T_{n}x,y)$ is bounded by applying polarization to the boundedness of $(T_{n}x,x)$.
Now we consider $S_{n}(y)=(y,T_{n}x)$, for each fixed $x$, Uniform Boundedness Theorem gives $\|S_{n}\|\leq M_{x}$, then $|(T_{n}x,y)|\leq M_{x}$ for any $y$ with $\|y\|\leq 1$. We set $y=T_{n}x/\|T_{n}x\|$ to get $\|T_{n}x\|\leq M_{x}$, once again Uniform Boundedness Theorem gives $\sup_{n}\|T_{n}\|<\infty$.
Best Answer
Counterexample: $T_n=-\frac1n{\rm Id}_H.$