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 $\varphi \in D(T^{**}) = D(T)$.

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.

No, this is never true if $\varphi \notin D(T)$.

This fact is probably well known but I hadn't seen it written down, so I wrote a version as Lemma 8.3 of these notes.

Proposition. Let $T$ be a closed and densely defined operator on a Hilbert space $H$, and fix $x \in H$. If there exists a sequence $x_n \in D(T)$ such that $x_n \to x$ (strongly or even weakly) and $\sup_n \|T x_n\| < \infty$, then $x \in D(T)$.

Proof. By Alaoglu's theorem, we can pass to a subsequence so that $T x_n$ converges weakly to some $y \in H$. Now let $v \in D(T^*)$ be arbitrary. We have $\langle T x_n, v \rangle = \langle x_n, T^*v \rangle$ for every $n$, and so passing to the limit, $\langle y, v \rangle = \langle x, T^*v \rangle$. This implies $x \in D(T^{**})$ and $T^{**}x=y$. But $T$ is closed and densely defined so $T^{**}=T$.

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 $\varphi \in D(T^{**}) = D(T)$.