Bound on the difference of two determinants

Since the most of related results seem to be published in articles devoted to the perturbation of determinant, let me rewrite your bound $$ \big\vert \, \det \left(\,A\,\right) - \det \left(\,B\,\right)\,\big\rvert \le n^{n+1} \left\|\,A-B\,\right\|_\infty \max \big\{ \left\|\,A\,\right\|^{n-1}_\infty, \,\left\|\,B\,\right\|^{n-1}_\infty \big\}\tag{1a}\label{1a} $$ under assumption $\,B = A + E$: $$ \big\vert \, \det \left(\,A\,\right) - \det \left(\,A+E\,\right)\,\big\rvert \le n^{n+1} \left\|\,E\,\right\|_\infty \max \big\{ \left\|\,A\,\right\|^{n-1}_\infty, \,\left\|\,A+E\,\right\|^{n-1}_\infty \big\}\tag{1b}\label{1b} $$

The best bound I could find is presented in this paper:

Absolute perturbation bounds (Theorem 2.12, p.768, [1]):

Let $A$ and $E$ be $n \times n$ complex matrices. Then $$ \big\vert \, \det \left(\,A\,\right) - \det \left(\,A+E\,\right)\,\big\rvert \le n \,\left\|\,E\,\right\|_2 \max \big\{ \left\|\,A\,\right\|_2, \,\left\|\,A+E\,\right\|_2 \big\}^{n-1} \tag{2}\label{2} $$

Relative perturbation bounds (Corollary 2.14, p.770, [1]):

Let $A$ and $E$ be $n \times n$ complex matrices. If $A$ is nonsingular, then $$ \frac{ \big\vert \, \det \left(\,A+E\,\right) - \det\left(\,A\,\right)\,\big\rvert }{ \big\vert\,\det \left(\,A\,\right)\,\big\rvert } \le \Big( \big\|\,A^{-1}\,\big\|_2 \big\|\,E\,\big\|_2\Big)^{n}-1 = \left(\kappa\,\frac{\left\|\,E\,\right\|_2}{\left\|\,A\,\right\|_2}\right)^{n}-1, \tag{3}\label{3} $$ where $ \kappa \equiv \big\|\,A\,\big\|_2 \,\big\|\,A^{-1}\,\big\|_2$.

The absolute bound result can be rewritten in terms of singular values of $A$.

Absolute perturbation bounds (Corollary 2.7, p.767, [1]): Let $A$ and $E$ be $n \times n$ complex matrices. Then $$\big\vert \, \det \left(\,A\,\right) - \det \left(\,A+E\,\right)\,\big\rvert \le\sum_{i=1}^{n} s_{n-i} \left\|E\right\|^i_2.\tag{2a}\label{2a}$$ If $\operatorname{rank}\left(A\right) = r$ for some $1 ≤ r ≤ n − 1$, then $$\big\vert \det \left(\,A+E\,\right)\big\rvert \le\left\|E\right\|^{n-r}_2 \sum_{i=1}^{r} s_{r-i} \left\|E\right\|^i_2,$$ where the $s_j$ are elementary symmetric functions in the $r$ largest singular values of $A$, $1 ≤ j ≤ r$. The bounds hold with equality for $E = \varepsilon UV_*$ with $ε > 0$, where $ A = UΣV_*$ is a SVD of $A$.

Let us compare, for example, $\eqref{1b}$ and $\eqref{2}$. Since $$ \frac{1}{\sqrt n}\left\|\,E\,\right\|_\infty \le \left\|\,E\,\right\|_2 \le \sqrt n\left\|\,E\,\right\|_\infty \implies \left\|\,E\,\right\|_\infty \ge n^{-\frac{1}{2}} \left\|\,E\,\right\|_2 , $$ we have $$ \begin{aligned} \eqref{1b} & = n^{n+1} \left\|\,E\,\right\|_\infty \max \big\{ \left\|\,A\,\right\|^{n-1}_\infty, \,\left\|\,A+E\,\right\|^{n-1}_\infty \big\} \ge \\ & \ge n^{n} \left( n^{-\frac{1}{2}}\left\|\,E\,\right\|_2\right) \max\big\{\left\|\,A\,\right\|_\infty,\,\left\|\,A+E\,\right\|_\infty\big\}^{n-1} = \\ & = n^{n+\frac{1}{2}} \left\|\,E\,\right\|_2 \max\big\{\left\|\,A\,\right\|_\infty,\,\left\|\,A+E\,\right\|_\infty\big\}^{n-1} \ge \\ & \ge n^{\frac{n}{2}+1} \left\|\,E\,\right\|_2 \max\Big\{\!\!\left( n^{-\frac{1}{2}}\left\|\,A\,\right\|_2\right), \, \left( n^{-\frac{1}{2}}\left\|\,A+E\,\right\|_2\right)\!\!\Big\}^{n-1} = \\ & = n^{\frac{n+3}{2}}\left\|\,E\,\right\|_2 n^{\frac{1-n}{2}} \max\Big\{\left\|\,A\,\right\|_2,\,\left\|\,A+E\,\right\|_2 \Big\}^{n-1} = \\ & = n^{2} \left\|\,E\,\right\|_2 \max\Big\{\left\|\,A\,\right\|_2,\,\left\|\,A+E\,\right\|_2 \Big\}^{n-1} \ge \\ & \ge n \left\|\,E\,\right\|_2 \max \big\{ \left\|\,A\,\right\|_2, \,\left\|\,A+E\,\right\|_2 \big\}^{n-1} = \eqref{2} \end{aligned} $$

Thus, we conclude that the estimate $\eqref{2}$ is sharper than $\eqref{1b}$.

Reference:

1. Ipsen, Ilse C. F., and Rizwana Rehman. "Perturbation Bounds for Determinants and Characteristic Polynomials." SIAM. J. Matrix Anal. & Appl. SIAM Journal on Matrix Analysis and Applications 30.2 (2008): 762-76. Web. 7 Aug. 2015.