Let $R$ be a Noetherian domain of dimension $\ge 1$. Let $\mathfrak{p}_i$, $i = 1, 2, ...$ be prime ideals of height one. Let $T = R[[X]]$ with $X$ is a indeterminate. For each $i \ge 1$ we set $\mathfrak{q}_i = \mathfrak{p}_i[[X]] = \mathfrak{p}_iT$ the extension of $\mathfrak{p}_i$. It is clearly that the height of $\mathfrak{q}_i$ is one for all $i$. Set $S = T \setminus \cup_{i \ge 1}\mathfrak{q}_i$.
Question: Is it true that $\dim T_S = 1$?
If you can use the Countable Prime Avoidance Lemma (Edit: that is, if you are in situations in which the Countable Prime Avoidance Lemma applies), I think the answer is yes. Because any prime ideal of $T_S$ comes from a prime ideal of $T$ that is contained in the union of the $\mathfrak{q}_i$'s. But then the Countable Prime Avoidance Lemma would imply that this prime ideal must be contained in one of the $\mathfrak{q}_i$'s. Hence $\dim T_S$ cannot be bigger than $1$.
The Countable Prime Avoidance Lemma will work, if for instance $R$ contains an uncountable number of elements $\{\mu_\lambda\}_\lambda$ such that $\mu_\lambda-\mu_\gamma$ is a unit in $R$ for every $\lambda\neq\gamma$. This condition will hold, if for example $R$ is local with $R/\mathfrak{m}$ uncountable.
A reference for this version of the Countable Prime Avoidance Lemma is page 242 of the book Cohen-Macaulay Representations by Graham Leuschke and Roger Wiegand, but I am sure you can find it at other places, too.
I will prove that $T_S$ satisfy Countable Prime Advoidance Lemma in the form mentioned in Mandi's answer (See the book of Graham Leuschke and Roger Wiegand)
First note that an element of $T = R[[X]]$ is a unit if its constant term is a unit in $R$. After localization such element is also a unit in $T_S$.
Since $X \notin \mathfrak{q}_i = \mathfrak{p}_i[[X]]$ we have $X$ contained in the multiplicative closed set $S$. Therefore $X$ is a unit in $T_S$.
Consider the following set $$B = \{u_{\mu} = b_0 + b_1X + \cdots + b_iX^i + \cdots | b_i = 0 \text{ or } 1, \text{ and } u_{\mu} \neq 0\}.$$ It is clearly that $B$ in a uncountable set of units in $T_S$.
Moreover, if $u$ and $v$ in $B$ we have $u - v$ (edit: $u-v$ nay not in $B$ but it is a unit) is a unit. Thus the assumption of Countable Prime Advoidance Lemma is satisfying.
The following application is a generalization of my question.
Corollary. Let $\{p_i\}_{i \ge 1}$ is a countable set of prime ideals with not containment relation. Let $T = R[[X]]$ and $S = T \setminus \cup_{i\ge 1}\mathfrak{p}_iT$. Then $R \to T_S$ is a flat extension and $$\text{Max}(T_S) = \{\mathfrak{p}_iT_S\}_{i \ge 1}.$$
Let me give a detailed account of Mahdi's argument.
First, we have to suppose (which is not totally clear in the question), that we consider only countably many primes $\mathfrak{p}_i$ of height $1$.
Second, we will suppose that $R/\mathfrak{m}$ is uncountable for every maximal ideal $\mathfrak{m}$ of $R$.
Then, $\dim(S^{-1}T)=1$.
Indeed, replacing $R$ by its localisation at a maximal ideal we can suppose that $R$ is in addition local. Then, in order to show that $\dim(S^{-1}T)=1$ we will show that every non-zero prime of $S^{-1}T$ has height $1$. So, let $\mathfrak{p}$ be a non-zero prime of $S^{-1}T$. Then, there exists a non-zero prime $\mathfrak{p}'$ of $T$ with $S\cap\mathfrak{p}'=\emptyset$ such that $\mathfrak{p}=S^{-1}\mathfrak{p}'$, and moreover $\mathfrak{p}$ and $\mathfrak{p}'$ have the same height. Thus it suffices to show that $\mathfrak{p}'$ has height $1$. The condition $S\cap\mathfrak{p}'=\emptyset$ is equivalent to $\mathfrak{p}'\subseteq\bigcup_{i\in\mathbb{N}}\mathfrak{q}_i$. Since $R$ is local with uncountable residue field, the same holds for $T$. Hence, countable prime avoidance holds in $T$, and so we see that there exists $i\in\mathbb{N}$ such that $\mathfrak{p}'\subseteq\mathfrak{q}_i$. As $\mathfrak{p}'$ is non-zero, $\mathfrak{q}_i$ has height $1$ and $T$ is a domain, it follows that $\mathfrak{p}'=\mathfrak{q}_i$ has height $1$ as desired.
(Using another variant of countable prime avoidance, also proven by Sharp and Vamos, gives the same conclusion under the hypothesis that $R$ is a complete local ring.)
