What is $\left|A - \frac{1}{\iota' A \iota} A\iota \iota' A\right|$? Let $A \in\mathbb{R}^{N \times N}$ be a symmetric matrix (of full rank) and $|A| := \sqrt{\sum\limits_{i=1}^N \sum\limits_{j=1}^N |a_{i,j}|}$ the Frobenius-Norm. What can we say about $\left|A - \frac{1}{\iota' A \iota} A\iota \iota' A\right|$ with $\iota$ being vectors of ones? Applying the obvious inequalities leads to: $$|A| \geq \sqrt{\iota' A \iota} \\\ \Rightarrow \left|A - \frac{1}{\iota' A \iota} A\iota \iota' A\right|\leq \left|A\right| + \frac{1}{|A|^2} |A\iota\iota'A| \leq |A| + N $$ However, I am afraid this inequality is not using most of the information available.

$A\in\mathbb{R}^{n\times n}$, symmetric and $e:=[1, 1, \ldots, 1]^{\top}$. Now,

\begin{equation} e^{\top}Ae=\sum_{i,j}A_{i,j}=:c \end{equation}

and,

\begin{align} Aee^{\top}A=\alpha\alpha^{\top} \end{align}

where $\alpha:=Ae$ denotes the vector of row (or column) sums of $A$.

So,

\begin{equation} \|A-\frac{1}{e^{\top}Ae}Aee^{\top}A\|_{F}=\|A-c\alpha\alpha^{\top}\|_{F} \end{equation}

One way to proceed can be to apply triangle inequality. For that, note that for a rank-1 symmetric matrix $xx^{\top}$, $\|xx^{\top}\|_{F}=\|x\|_{2}^{2}$. This is because for any matrix, the Frobenius norm is the $2$-norm of the vector of singular values, and $xx^{\top}$ has just one non-zero eigenvalue, namely, $\|x\|_{2}^{2}$. This gives

\begin{equation} \|A-\frac{1}{e^{\top}Ae}Aee^{\top}A\|_{F}\le \|A\|_{F}+\frac{1}{c}\|\alpha\|_{2}^{2}. \end{equation}

Is this the kind of characterization you are looking for?