Let $A\otimes A$ denote the Kronecker product. Suppose $\|A - B\|_1 = \varepsilon$, where $\|\cdot \|_1$ is the nuclear norm a defined by $\|X\|_1 = \text{Tr}(\sqrt{X^\dagger X})$ and $X^\dagger$ is the transpose conjugate.
What can be said about $\|A\otimes A - B\otimes B\|_1$ or more generally about $\|A^{\otimes n} - B^{\otimes n}\|_1$ in terms of $\varepsilon$?
One bound is as follows. We have $$ \|A \otimes A - B \otimes B\|_1 \leq \\ \|A \otimes A - A \otimes B\|_1 + \|A \otimes B - B \otimes B\|_1 \leq\\ \|A \otimes (A - B)\|_1 + \|(A - B) \otimes B\|_1 = \\ \|A\|_1 \cdot \|A - B\|_1 + \|A - B\|_1 \cdot \|B\|_1 =\\ (\|A\|_1 + \|B\|_1)\|A - B\|_1\leq\\ 2\max\{\|A\|_1,\|B\|_1\} \cdot \|A - B\|_1 = \\ 2\max\{\|A\|_1,\|B\|_1\} \cdot \varepsilon $$ Applying similar logic in the general case leads to the inequality $$ \|A^{ \otimes n} - B^{\otimes n}\|_1 \leq n\cdot [\max\{\|A\|_1,\|B\|_1\}]^{n-1} \cdot \varepsilon $$