Let $\Omega\subset \mathbb{R}^n$ be a bounded smooth domain and $\{p_m^*\}\subset L^2(\Omega)$ a sequence. I know that for every $m\in \mathbb{N}$ the element $v(p_m^*)\in H_0^1(\Omega)^n$ satisfies $$((v(p_m^*),w)) = (f,w) + (p_m^*, \text{div }w) \quad \forall w\in H_0^1(\Omega)^n,$$ where $((u,v))= \sum_{1\leq i,j \leq n}(D_iu_j, D_iv_j)$.
I also know that the sequence $\{v(p_m^*)\}$ is bounded in $H_0^1(\Omega)^n$.
QUESTION: How to show that $\{\nabla p_m^*\}$ is bounded in $H^{-1}(\Omega)^n$, which is the dual of $H_0^1(\Omega)^n$?
By definition, the weak gradient satisfies $$ \int_\Omega \nabla p_m \cdot \phi dx = -\int_\Omega p_m \ div(\phi) dx $$ for all $\phi\in C_c^\infty(\Omega)^n$. Using your equation, this is equivalent to $$ \int_\Omega \nabla p_m \cdot \phi dx = (f,\phi) - (( \nu(p_m),\phi)). $$ Hence it holds $$ \left | \int_\Omega \nabla p_m \cdot \phi dx \right| \le \|f\|_{H^{-1}} \|\phi\|_{H^1} + \|\nu(p_m)\|_{H^1})\|\phi\|_{H^1}. $$ This shows that the functionals $$ \phi \mapsto \int_\Omega \nabla p_m \cdot \phi dx $$ are functionals on $H^1_0(\Omega)$, and bounded in $H^{-1}$, which implies that the weak gradients of $p_m$ are bounded in $H^{-1}$.