No, this only holds when $\varphi \in D(T)$.

It's enough to assume that $T$ is closed and densely defined, so that $T^{**} =T$.  Let $\psi \in D(T^*)$ be arbitrary.  If the hypothesis holds then we have
$$ |\langle \varphi, T^* \psi \rangle| = \lim_{n \to \infty} |\langle \varphi_n, T^* \psi \rangle| = \lim_{n \to \infty} |\langle T \varphi_n, \psi \rangle| \le k(\varphi) \|\psi\|$$
which is exactly the definition of $\varphi \in D(T^{**}) = D(T)$.  One can also see that in fact $T \varphi_n \to T \varphi$ weakly.



Note the argument still goes through even if we only assume that $\varphi_n \to \varphi$ weakly instead of strongly.