Is a domain all of whose localizations are noetherian itself noetherian ?

I had the exact same question not too long ago. Apparently if you drop the noetherian precondition in Neukirch's definition of "Dedekind domain" then you get what some people call an "almost Dedekind domain". There are indeed examples of almost Dedekind domains that aren't Dedekind (i.e. aren't noetherian). The first of these was given by Nakano (J. Sci. Hiroshima Univ. Ser. A. 16, 425–439 (1953)): take the integral closure of $\mathbb Z$ in the field obtained by adjoining to $\mathbb Q$ the $p$th roots of unity for all primes $p$.

No, this isn't true. In the paper [Heinzer, Ohm: Locally Noetherian Commutative Rings] the authors (who also mention Nakano's example from Faisal's answer which they call "quite involved" ) construct two counter-examples: see Examples 2.2, 2.3.

Their example 2.3 shows moreover that $D$ doesn't have to be noetherian, even if the space $Spec(D)$ and all $D_P$ are noetherian.

All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=\mathbb{Z} \setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.