I wouldn't be so quick to discount the value of going through the list one by one. For this post, let's stipulate $D < 0$ throughout.
You've noticed $-2$ is the only even value on there, right? If $D$ is even, then $D = (\sqrt D)^2$, which is obvious enough. But if $D$ is even and composite, it means that $N(z) = 2$ for $z \in \mathcal{O}_{\mathbb{Q}(\sqrt D)}$ is impossible. So then $2$ is irreducible, yet $D = 2 \times x = (\sqrt D)^2$, where $x \in \mathcal{O}_{\mathbb{Q}(\sqrt D)}$ also.
You've also noticed that $-5$ is not on the list either. Neither is $-13$, $-17$, $-29$, etc. What these numbers have in common, besides being odd, is that they are congruent to $3 \bmod 4$ (remember that congruence gets "flipped" for negative numbers, so $-3 \equiv 1 \bmod 4$, not $3 \bmod 4$).
So, if $D \equiv 3 \bmod 4$, then $N(1 + \sqrt D) = -D + 1$, which is even. But in this domain, it turns out that $N(z) = 2$ is also impossible. Which means that $-D + 1$ has at least two distinct factorizations. Thus $D = -5$ gives us the classic example $6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5})$.
Now let's say $D \equiv 1 \bmod 4$ instead. Then it's still the case that $-D + 1 = (1 - \sqrt{D})(1 + \sqrt{D})$, but... $$\frac{1 - \sqrt{D}}{2}, \frac{1 + \sqrt{D}}{2}$$ are also algebraic integers, both with minimal polynomial $$x^2 - x + \frac{-D + 1}{4},$$ e.g., $$\frac{1 - \sqrt{-43}}{2}, \frac{1 + \sqrt{-43}}{2}$$ both have the polynomial $x^2 - x + 11$.
Therefore, the full factorization of $44$ in this domain is not $(1 - \sqrt{-43})(1 + \sqrt{-43})$ but $$2^2 \left(\frac{1 - \sqrt{-43}}{2}\right) \left(\frac{1 + \sqrt{-43}}{2}\right).$$
Why this doesn't work out for $D \leq -167$ is quite a bit more involved, maybe someone else will address that.
I'll give an "elementary" answer to this question based on the comments in the question:
Let $\alpha\in \mathcal{O}$ with $N(\alpha)=p$ a prime in $\mathbb{Z}$. Write $\mathcal{O}=\mathbb{Z}\oplus \lambda_d\mathbb{Z}$ and consider the abelian group homomorphism $\varphi:\mathbb{Z}\oplus \lambda_d\mathbb{Z}\to \mathbb{Z}\oplus \lambda_d\mathbb{Z}$ given by $\varphi(x)=\alpha x$. In the base $\{1,\lambda_d\}$ we can represent $\phi$ by a matrix $[\varphi]$, and a simple computation (considering $d\equiv 1$ (mod $4$) and $d\equiv 2,3$ (mod $4$) separately) gives
$$
\det([\varphi]) = N(\alpha).
$$
By using the Smith normal form, there are matrices $P,Q\in GL(n,\mathbb{Z})$ such that $\det(P)=\det(Q)=1$ and
$$
[\varphi] = P\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix}Q,
$$
and then
$$
\varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z}) \cong (\mathbb{Z}/d_1\mathbb{Z})\oplus (\mathbb{Z}/d_2\mathbb{Z}),
$$
so we have that $|\varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z})|=d_1d_2=\det([\varphi])=N(\alpha)=p$. But $\varphi(\mathbb{Z}\oplus \lambda_d\mathbb{Z})=(\alpha)$ is the principal ideal generated by $\alpha$ in $\mathcal{O}$, so, by the first isomorphism theorem, $|\mathcal{O}/(\alpha)|=p$. Then we can conlude (see for example this answer) that $\mathcal{O}/(\alpha)$ is isomorphic to the field $\mathbb{F}_p$, so $(\alpha)$ is a prime ideal and $\alpha$ is a prime elment in $\mathcal{O}$.
Best Answer
If I'm understanding correctly, your question is whether it's possible for $p$ to be a positive prime number in $\mathbb Z$ and also prime in each of $\mathbb Z[i]$, $\mathbb Z[\sqrt{-2}]$, $\mathbb Z[\omega]$, $\mathcal O_{\mathbb Q(\sqrt{-7})}$, $\mathcal O_{\mathbb Q(\sqrt{-11})}$, $\mathcal O_{\mathbb Q(\sqrt{-19})}$, $\mathcal O_{\mathbb Q(\sqrt{-43})}$, $\mathcal O_{\mathbb Q(\sqrt{-67})}$ and $\mathcal O_{\mathbb Q(\sqrt{-163})}$? And if so, what is the smallest such $p$?
In a comment yesterday, I wrote I didn't think such a prime exists. I had failed to notice your earlier comment about 3167. This suggests that these primes do exist but are spaced far apart.
Obviously $p$ must not be one of 2, 3, 7, 11, 19, 43, 67 or 163, which still leaves an infinitude of primes.
But the criterion for $\mathbb Z[i]$ gives us a way to discard "half" the primes: we're looking for $p \equiv 3 \pmod 4$, though obviously $p \neq 3$ itself.
I'm not sure why you have 0, 2, 4, 6 in the criterion for $\mathbb Z[\sqrt{-2}]$, there should be no even values in that one. This leaves $p \equiv 5, 7 \pmod 8$, but we can cross off 5 since $5 \pmod 8 \equiv 1 \pmod 4$, further narrowing it down to $7 \pmod 8$.
In order to combine this with the criterion for $\mathbb Z[\omega]$, we need to broaden our modulus to 24. Hence $p \equiv 7, 15, 23 \pmod{24}$. But clearly $15 \pmod{24}$ can't be prime. And $7 \pmod{24}$ can't be prime in $\mathbb Z[\omega]$ since it's equivalent to $1 \pmod 3$. That leaves us $p \equiv 23 \pmod{24}$.
Next we move to the Kleinian integers — hmm, I don't think I had ever come across that term before seeing your question. We need to broaden our modulus out again, to 168 this time, so we have $p \equiv 23, 47, 71, 95, 119, 143, 167 \pmod{168}$.
It shouldn't bother us too much at this point that $95 = 5 \times 19$, since $\gcd(95, 168) = 1$. Likewise with $143 = 11 \times 13$. It is far more important that $119 = 7 \times 17$, so for that reason we discard 119. So our possibilities are $p \equiv 23, 47, 71, 95, 143, 167 \pmod{168}$.
However, $$\left(\frac{-7}{23}\right) = (-7)^{11} \pmod{23} = 1$$ and indeed $(4 - \sqrt{-7})(4 + \sqrt{-7}) = 23$. $p = 47$ is still in the running since $$\left(\frac{-7}{47}\right) = (-7)^{23} \pmod{47} = -1.$$ But 71 is out, with $(8 - \sqrt{-7})(8 + \sqrt{-7}) = 71$.
For the sake of the Legendre symbol, let's substitute 95 with 263. We quickly find that $(16 - \sqrt{-7})(16 + \sqrt{-7}) = 263$. Replacing 143 with 311, we find it's still in the running. So is 167 (as a residue class, not as a number itself).
So our list of possibilities is now $p \equiv 47, 143, 167 \pmod{168}$, and we have to broaden our modulus out once again to 1848. Taking a shortcut is looking very good right about now. But let's soldier on with 1848.
The possibilites explode, so instead of going through all of them one by one, I'm gonna try to take more of these on with Mathematica's help (you can use Wolfram Alpha, to some extent, if you don't have Mathematica on your computer). So
JacobiSymbol[-11, 47 + 168Range[0, 10]]gives us the sequence $1, -1, 1, 1, 1, -1, -1, -1, 1, -1, 0$.But that second $-1$ corresponds to 215, so we substitute 2063 and see it's still in the running...
That's where I left it last night. This morning, I decided it would make more sense to just do a brute force search in Mathematica and see if anything comes up.
And then there was your gentle reminder about 3167, which checks out with the Legendre symbol in Mathematica (directly, or indirectly through Wolfram Alpha) as well as on the command line with a little library I got off GitHub. So I ran the the brute force search in Mathematica among the first two thousand primes and it gave me 3167, 8543, 14423, which are all $143 \pmod{168}$.
But there are plenty of other primes congruent to $143 \pmod{168}$, why don't they also stay inert in the nine quadratic imaginary UFDs? 311, for example. Indeed,
Maybe if we had broadened our modulus all the way out to 16488700536, we might find something along the lines of that only primes congruent to 3167 and maybe a few other values modulo 16488700536 can also be prime in all nine quadratic imaginary UFDs.
P.S. If you're interested, 15073 is composite in all these rings.