= Solution
First $N(A)$ is finite: by <Cauchy-Schwarz>, $\|Ax\|_2\le(\sum_{ij}a_{ij}^2)^{1/2}\|x\|_2$. It is nonnegative and absolutely homogeneous. If $N(A)=0$, every $Ae_j=0$, so every column is zero and $A=0$; the converse is immediate. The triangle inequality for the <Euclidean norm> gives $N(A+B)\le N(A)+N(B)$ after taking suprema. Thus \b[$N$ is a <norm>].
For any vector $x$, rescaling gives $\|Ax\|_2\le N(A)\|x\|_2$. Applying this twice gives
$$
\boxed{N(AB)\le N(A)N(B).}
$$
Furthermore $|a_{ij}|\le\|Ae_j\|_2\le N(A)$, proving the requested strict entry bound when $N(A)<\varepsilon$.
For a <polynomial> in <matrix> entries, expand each monomial at $A+H$. The terms involving exactly one entry of $H$ give a <linear map> $Df(A)[H]$, with <coefficients> <polynomial> in the entries of $A$. Every remaining term has at least two entries of $H$ and is $O(N(H)^2)$ for $N(H)\le1$, because $|h_{ij}|\le N(H)$. This proves differentiability. Its <derivative> is continuous as an operator-valued function: each <coefficient> is continuous, and for $N(H)\le1$ every $h_{ij}$ is bounded by one, so the operator-<norm> difference is bounded by the sum of the <coefficient> differences. Coordinate <continuity> follows from the same entry bound. Hence \b[every such <polynomial> function is continuously differentiable].
In the Leibniz expansion of $\det(I+H)$, the identity permutation supplies $1+\sum_i h_{ii}$ plus terms of degree at least two. Any nonidentity permutation moves at least two indices, so each of its terms has at least two $H$ entries. Consequently
$$
\det(I+H)=1+\operatorname{tr}H+O(N(H)^2),\qquad
\boxed{d'(I)[H]=\operatorname{tr}H.}
$$
If $A$ is invertible, multiplicativity of the <determinant> and the preceding expansion give
$$
\det(A+H)=\det A\det(I+A^{-1}H)
=\det A+\det A\operatorname{tr}(A^{-1}H)+O(N(H)^2).
$$
The adjugate identity gives $\operatorname{adj}A=\det(A)A^{-1}$. Thus $d'(A)[H]=\operatorname{tr}((\operatorname{adj}A)H)$ for invertible $A$.
For singular $A$, take invertible $A_r\to A$, using the permitted density result. Both $d'$ and the adjugate entries are continuous <polynomial> expressions, so taking limits for every fixed $H$ proves the <derivative of the determinant> formula on all <matrices>:
$$
\boxed{d'(A)[H]=\operatorname{tr}((\operatorname{adj}A)H).}
$$
Here the standard adjugate is the transpose of the cofactor array, as required by $(\operatorname{adj}A)A=(\det A)I$. In components this <derivative> is $\sum_{ij}\operatorname{cof}_{ij}(A)h_{ij}$, also valid at singular <matrices>.
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