Bound on the difference of two determinants Let $A$ and $B$ be two real, $n\times n$ matrices. Using Hadamard's inequality, it is not hard to show that
$$
\left|\det A - \det B \right| \leq \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}.
$$
Where $\|A\|_2=\sqrt{\sum_{i,j}a_{ij}^2}$.  From this, I can derive a sup bound, for example 
$$
\left|\det A - \det B \right| \leq n^{n+1} \|A-B\|_{\infty} \max (\|A\|_{\infty}^{n-1},\|B\|_{\infty}^{n-1}).
$$
Where $\|A\|_\infty=\sup_{i,j}|a_{ij}|$. 
The constant $n^{n+1}$ is not the best bound possible : any reference (or proof) for a better (or the best) one? I show below that one can obtain $n^2(n-1)^{n-1}$, but that isn't much better.
I just tried  $10^5$ random matrices on Maple and obtained a maximal constant (much) smaller  than one : this is not a proof, but it looks like there is room for improvement nevertheless.

Just for completeness (and in case someone sees a factor I missed), to get the first bound, writing $A=[A_1,\ldots,A_n]$ in terms of its column vectors, an expansion shows
\begin{eqnarray*}
\det A &=& \det (A_1 -B_1,A_2,\ldots,A_n) + \det (B_1,A_2,\ldots,A_n) \\
  &=& \sum_{j=1}^n \det (B_1,\ldots, B_{j-1}, A_j -B_j,A_{j+1},\ldots,A_n) \\
&& + \det B, 
\end{eqnarray*}
Thus by Hadamard's inequality,
\begin{eqnarray*}
\det A -\det B &\leq&  \sum_{j=1}^n \prod_{i=1}^{j-1} \|B_i\|_{2}\prod_{i=j+1}^{n} \|A_i\|_{2} \|A_j-B_j\|_{2} \\
&\leq&  \|A-B\|_{2} \sum_{j=1}^n \|B\|^{j-1}_{2}\|A\|^{n-j}_{2} \\ 
&=& \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}.
\end{eqnarray*}
The second bound is just that $x^n -y^n\leq n \max(|x|^{n-1},|y|^{n-1}) |x-y|$ and $\|A\|_2 \leq n\|A\|_\infty$.

Another approach is calculus, namely, to write that
$\det B - \det A = f(1)-f(0)$, with $f(t)=\det(A + t(B-A))$. 
By the mean value theorem $f(1)-f(0)\leq \max |f^\prime (t)|$. 
We can compute that $$f^\prime(t) =  {\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)$$ 
(if I did not mess up, using the formula for the differential of a determinant, where Cofm means the matrix of Cofactors). 
Then, it should deliver something better, if there is a nice way to bound it. The simplest thing is to use Cauchy-Schwarz, namely
$$
\left|{\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)\right|\leq \|B-A\|_{2} \|{\rm Cofm}(A +t(B-A))\|_{2}
$$
and then, by lack of a better idea, 
$$
\|{\rm Cofm}(A +t(B-A))\|_{2}\leq n \max_{ij} |{\rm Cof}_{i,j}(A +t(B-A))|,
$$
and brutally, $|{\rm Cof}_{i,j}(A +t(B-A))|\leq ((n-1) \max(\|A\|_\infty,\|B\|_\infty))^{n-1}$ gives a slightly better constant, namely
$$
n^2(n-1)^{n-1} <n^{n+1}
$$
but that still seems a very rough way to bound a determinant, as it is never sharp, since to attain this bound all coefficients should be equal, and therefore the cofactor would be zero. They are of the same order in $n$, and I suspect this order is wrong.
 A: 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:


*

*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.
