Let $A$ be a symmetric matrix in $\mathbb R^n$ such that $A$ is positive definite and hence satisfies $0< \lambda \le A \le \Lambda < \infty.$


Let $T$ be a densely defined and closed operator from some Hilbert space $H$ into $H^n$. It is a classical theorem by John von Neumann that $T^\*T$ is self-adjoint with domain $D(T^\*T).$


I wonder whether it is true that for some $C>0$


$$\Vert \langle AT,T \rangle x \Vert \le C (\Vert T^\*T x \Vert + \Vert x \Vert) \text{ and all } x \in D(T^\*T).$$


Similarly, it seems natural to ask whether we also have that 


$$\Vert T^\*T x \Vert \le C (\Vert \langle AT,T \rangle x \Vert + \Vert x \Vert) \text{ and all } x \in D(\langle AT,T \rangle)?$$



Elliptic estimates for self-adjoint operators
As I understand your question now, after edit, in dimension two you ask whether
$$
 \|T\_2^\* T\_1 x\| \leqslant C(\|T\_1^\* T\_1 x + T\_2^\* T\_2 x\| + \|x\|)
$$
whenever $T\_1$, $T\_2$ are densely defined closed operators; $C$ can depend on $T\_1$ and $T\_2$.


This need not be the case. Let $T\_1$ be the identity operator acting on $\ell^2$, and let $T\_2$ be given by the matrix
$$ T\_2 = \pmatrix{0&1&0&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&2&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&0&0&3&\cdots\\0&0&0&0&0&0&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} . $$
Then
$$ T\_1^\* T\_1 + T\_2^\* T\_2= \pmatrix{1&0&0&0&0&0&\cdots\\0&1 + 1&0&0&0&0&\cdots\\0&0&1&0&0&0&\cdots\\0&0&0&1 + 4&0&0&\cdots\\0&0&0&0&1&0&\cdots\\0&0&0&0&0&1 + 9&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} , $$
and
$$ T\_2^\* T\_1 = T\_2^\* = \pmatrix{0&0&0&0&0&0&\cdots\\1&0&0&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&2&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&0&3&0&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} . $$
In particular, if $e\_k$ is the $k$-th vector of the canonical basis of $\ell^2$, then $$\|T\_2^\* T\_1 e\_{2n-1}\| = \|n e\_{2n}\| = n,$$ but $$\|T\_1^\* T\_1 e\_{2n-1} + T\_2^\* T\_2 e\_{2n-1}\| = \|e\_{2n-1} + 0\| = 1 .$$
Therefore a constant $C$ with the desired property does not exist.


