Is every commutative ring a limit of noetherian rings?
The answer is no to all questions except 4.
Negative answers to 1,2 and 3:
It is easy to construct a ring $A$ with an element $a$ satisfying:
(i) $a≠0$,
(ii) $a$ is nilpotent,
(iii) for each $n≥1$ there is $y_n\in A$ such that $a=y_n^n$.
For every morphism $\varphi:A\to C$, the image of $a$ inherits properties (ii) and (iii), hence it is zero if $C$ is noetherian (observe that the radical of $C$ is finitely generated, hence nilpotent). In other words, any morphism from $A$ to a noetherian ring $C$ factors through $B:=A/(a)$, and of course the same holds if $C$ is a limit of noetherian rings. This proves that $A$ is not such a limit (Question 1) and the natural morphism $A\to B$ provides a negative answer to Question 2.
For Question 3, consider the following special case: let $k$ be a nonzero noetherian ring and take $$A:=k[X_1,X_2,\dots]/(X_1, X_{mn}^m-X_n)_{m,n≥1}.$$ If $x_n$ denotes the class of $X_n$, we easily check that $x_n^n=0$ for all $n$, $x_n≠0$ if $n≥2$, and each $x_n$ is an $m$-th power for all $m$. If $\varphi:A\to C$ is a morphism with $C$ noetherian, the above argument (with $a=$any $x_n$) shows that $\varphi$ factors through $A/(x_n)_{n≥1}\cong k$. Now if $\Lambda$ is the ordered set of finitely generated $k$-subalgebras of $A$, then of course $A$ is the colimit of $\Lambda$ in CRing, but the above shows that there is a colimit in Noeth, which is $k$.
Positive answer to 4: $\mathbb{Z}[x_1,x_2,\dots]$ is a Krull domain (even a UFD), hence an intersection of discrete valuation rings inside its fraction field.
Negative answer to 5: Given a ring $R$, let us call a noetherian $R$-algebra $S$ a noetherian hull of $R$ if it is an initial object of the category of noetherian $R$-algebras.
Proposition. If $R$ has a noetherian hull $S$, the natural map $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ is surjective. In particular, $\mathrm{Spec}(R)$ is a noetherian space.
Proof: for each $p\in \mathrm{Spec}(R)$ its residue field $\kappa(p)$ (i.e. $\mathrm{Frac}(R/p)$) is a noetherian $R$-algebra, hence $R\to\kappa(p)$ factors through $S$. QED
(Remark: it is easy to see that $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ is in fact bijective, with trivial residue field extensions).
Now, if $A$ and $B$ are two noetherian rings, a coproduct of $A$ and $B$ in Noeth is the same thing as a noetherian hull of $A\otimes B$: this is clear from the universal properties. Thus, to answer Question 5 negatively, it suffices to find two noetherian rings $A$, $B$ such that $\mathrm{Spec}(A\otimes B)$ is not a noetherian space. There are plenty of examples, for instance $A=B=\overline{\mathbb{Q}}$.
Note: the same example was given by François Brunault in his answer, with a different argument. Since I have edited my answer several times, I am not sure who came first!
Binary coproducts do not always exist in $\textrm{Noeth}$.
Assume that the coproduct $C=\overline{\mathbb{Q}} \sqcup \overline{\mathbb{Q}}$ exists in $\textrm{Noeth}$. Letting $A=\overline{\mathbb{Q}} \otimes_{\mathbb{Z}} \overline{\mathbb{Q}}$, we then have a canonical ring map $\varphi : A \to C$. Now the idea is to consider some kind of completion of $A$, similar to the one you suggest in the last paragraph of your question: for every $\sigma \in G=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, there is a noetherian quotient $\mu_\sigma : A \to \overline{\mathbb{Q}}$ given by $\mu_\sigma(x \otimes y)=x \sigma(y)$. Moreover, the resulting morphism $i : A \to \prod_G \overline{\mathbb{Q}}$ is injective (this can be checked by restricting to $K \otimes K$ where $K$ is any finite Galois extension of $\mathbb{Q}$). Applying the coproduct property in $\textrm{Noeth}$, each $\mu_\sigma$ factors through $C$, so that $i$ factors through $\varphi$. In particular, the map $\varphi$ is injective, and we may identify $A$ with a subring of $C$.
Since $A$ is integral over $\overline{\mathbb{Q}}$, it has Krull dimension $0$. In particular, every maximal ideal $\mathfrak{m}_\sigma=\ker \mu_\sigma$ is a minimal prime ideal of $A$. By a result in Bourbaki, Commutative algebra (Chap. 2, Sect. 2.6, Prop. 16), the ideal $\mathfrak{m}_\sigma$ lies below a minimal prime ideal $\mathfrak{p}_\sigma$ of $C$. So we have constructed infinitely many minimal prime ideals of $C$, which is absurd because $C$ is noetherian.