Short version
$$\left\| \pmatrix{a&0&0\\ b&0&0 \\ c&0&0} \right\|^2 = a^2+b^2+c^2 \ \text{ and } \ \left\| \pmatrix{a&a&0\\ b&-c&0 \\ c&b&0} \right\|^2 = 2a^2+b^2+c^2.$$
These two forms together cover every natural number by Theorems I and V in this paper of L.E. Dickson.
Long version
Recall that $\Vert A \Vert^2$ is just the largest eigenvalue of $A^*A$. Take $a_i,b_i \in \mathbb{Z}$ and $$A=\pmatrix{a_1&b_1&0\\ a_2&b_2&0 \\ a_3&b_3&0}$$
Then $$ A^*A = \pmatrix{a_1^2 + a_2^2 + a_3^2&a_1b_1 + a_2b_2 + a_3b_3&0\\ a_1b_1 + a_2b_2 + a_3b_3&b_1^2 + b_2^2 + b_3^2&0 \\ 0&0&0}$$
We deduce its characteristic polynomial and the largest root. It follows that $$\Vert A \Vert^2 = \frac{1}{2} \left( \sum_{i=1}^3 (a_i^2 +b_i^2) + \sqrt{\left(\sum_{i=1}^3 (a_i^2 +b_i^2)\right)^2 -4\sum_{i<j}(a_ib_j-a_jb_i)^2} \right)$$
Let $u=\pmatrix{a_1\\ a_2 \\ a_3}$, $v=\pmatrix{b_1\\ b_2 \\ b_3}$, and $u\times v$ be their cross product. Then, observe that
$$\Vert A \Vert^2 = \frac{1}{2} \left( \Vert u \Vert^2 + \Vert v
\Vert^2 + \sqrt{\left(\Vert u \Vert^2 + \Vert v \Vert^2\right)^2 -4 \Vert u
\times v\Vert^2} \right)$$
Recall that $ \Vert u \times v \Vert^2 + (u \cdot v)^2 = \Vert u \Vert^2\Vert v \Vert^2$, with $u \cdot v$ the dot product. Then
$$\Vert A \Vert^2 = \frac{1}{2} \left( \Vert u \Vert^2 + \Vert v
\Vert^2 + \sqrt{\left(\Vert u \Vert^2 - \Vert v \Vert^2\right)^2 +4 (u
\cdot v)^2} \right)$$
Assume that $\Vert u \Vert = \Vert v \Vert$. Then $$ \Vert A \Vert^2 = \Vert u \Vert^2+ \vert u \cdot v \vert.$$
For any $u= \pmatrix{a\\ b \\ c} \in \mathbb{Z}^3$, take $v= \pmatrix{a\\ -c \\ b}$. Then $$ \Vert A \Vert^2 = 2a^2+b^2+c^2.$$
By Theorem V in this paper of L.E. Dickson, the above form represents every natural numbers not of the form $2^{2n+1}(8m+7)$. But this last is in $E_3$ by Legendre's three-square theorem, and we already know that $E_3 \subset F_3$. The result follows. $\square$
Bonus problem: Find a proof with $A \in M_3(\mathbb{N})$.
For fun: a classification of the natural numbers by angles
Recall that $\Vert u \times v\Vert^2 = \Vert u \Vert^2 \Vert v \Vert^2 \sin^2(u,v)$. Let's recall the above matrix $A$ as $A_{u,v}$. Consider the angle $$\alpha(n):=\min_{u,v \in \mathbb{Z}^3}\{\text{angle}(u,v) \in [0,2\pi) \text{ such that } \Vert A_{u,v} \Vert^2 = n \}.$$
Theorem: $\alpha(n) = 0$ if and only if $n \in E_2E_3$.
proof: Note that $\alpha(n) = 0$ iff $\exists u,v \in \mathbb{Z}^3$ with $\Vert A_{u,v} \Vert^2 = n$ and $u \times v = 0$ (i.e. collinear), iff $\exists r \in \frac{1}{\gcd(u)}\mathbb{Z}$ such that $v=ru$, with $\gcd(u)$ the greatest common divisor of $u_1, u_2$ and $u_3$. Then $$\Vert A \Vert^2= (r^2+1)\Vert u \Vert^2.$$
For any $u'= \pmatrix{a\\ b \\ c} \in \mathbb{Z}^3$ and any $s,t \in \mathbb{Z}$, assume that $u=su'$ (so that $s | \gcd(u)$) and $r=t/s$. Then $$\Vert A \Vert^2= (t^2+s^2)\Vert u' \Vert^2 = (t^2+s^2)(a^2+b^2+c^2).$$
The result follows. $\square$
By above material with $\vert u \cdot v \vert^2 = \Vert u \Vert^2 \Vert v \Vert^2 \cos^2(u,v)$, we have:
Lemma: $\alpha(2a^2+b^2+c^2) \le \arccos(\frac{a^2}{a^2+b^2+c^2})$.
Then, $\alpha(7) \in (0,\theta]$, with $\theta = \arccos(1/6) \simeq 1.403348 \text{ rad} \simeq 80.4°$