You can of course define $T$ by this limit, but this operator need not be closed. Take for example $H=\ell^2$ and $T_n$ as multiplication by $(0,\ldots, 0, n, n+1,\ldots)$ on its natural domain $D$. Note that $D$ is independent of $n$. We have $\langle x, T_n y\rangle\to 0$, but the zero operator is not closed on $D$.

These operators do not satisfy your extra assumption that the range is also contained in $D$. This assumption is quite restrictive and perhaps a bit unnatural. For example, it implies that $\sigma(T_n)=\mathbb C$. We can, however, modify the above example and obtain this property also. To do this, start out with multiplication by $(0,2,0,4,0,6,\ldots)$, followed up by a shift. Then take $T_n$ again as a cut off (at $n$) version of this. So $T_n x = (0, \ldots , 0, 2x_n, 0, 4x_{n+2}, 0,\ldots)$, with the non-zero entries sitting in the odd slots now.

Even if you can define $T$ by this limit, this operator need not be closed. Take for example $H=\ell^2$ and $T_n$ as multiplication by $(0,\ldots, 0, n, n+1,\ldots)$ on its natural domain $D$. Note that $D$ is independent of $n$. We have $\langle x, T_n y\rangle\to 0$, but the zero operator is not closed on $D$.

These operators do not satisfy your extra assumption that the range is also contained in $D$. This assumption is quite restrictive and perhaps a bit unnatural. For example, it implies that $\sigma(T_n)=\mathbb C$. We can, however, modify the above example and obtain this property also. To do this, start out with multiplication by $(0,2,0,4,0,6,\ldots)$, followed up by a shift. Then take $T_n$ again as a cut off (at $n$) version of this. So $T_n x = (0, \ldots , 0, 2x_n, 0, 4x_{n+2}, 0,\ldots)$, with the non-zero entries sitting in the odd slots now.