Let $\textbf{A}\in(H^1(\Omega))^3$, where $\Omega\subset\mathbb{R}^3$ is a bounded convex domain with its boundary $\partial\Omega$. Now we know, on $\partial\Omega$, $$\textbf{A}\times\textbf{n}=\textbf{m},$$ $$\textbf{A}\cdot\textbf{n}=0,$$ where $\textbf{n}$ is unit outer normal vector. Is the following formulation right? $$\|\textbf{A}\|_{(H^\frac{1}{2}(\partial\Omega))^3}=\|\textbf{m}\|_{(H^\frac{1}{2}(\partial\Omega))^3}$$
Plausible reason: the assumption $\textbf{A}\cdot\textbf{n}=0$ implies $|\textbf{A}\times\textbf{n}| = |\textbf{A}|\ |\textbf{n}|= |\textbf{A}|$.
No, this is not correct. The orthogonality ensures $|\mathbf A| = |\mathbf m|$, hence the norms of the magnitudes of these vector fields are the same.
But unlike the Lebesgue norms, the Sobolev norms are affected by the direction of the vector field, since they takes derivatives into account. For example, a unit vector field on a unit ball can have arbitrarily large Sobolev norm if its direction changes rapidly. In your case, the direction of $\mathbf m$ is affected by the geometry of the boundary, which will contribute to the norm $\|\mathbf m\|_{H^{1/2}}$.