Let $R$ be a Noetherian domain with fraction field $K\ne R$. Let $\overline R$ be the integral closure of $R$ in $K$.
Let us call that a subring of the fraction field $R\subseteq R^{fn}\subseteq K$ is the finite closure of $R$ if $R^{fn}$ is a finitely generated $R$-module and for any ring $R\subseteq S\subseteq K$ that is also a finitely generated $R$-module, we must have $S\subseteq R^{fn}$ . Note that, this finite closure, if exists, is unique and $R^{fn}\subseteq \overline R$ .
So my first question: Does $R^{fn}$ always exists ? What if we assume the ring is local ?
I believe the answer to the above question to be negative. Now notice that if $\overline R$ is a finitely generated $R$-module i.e. if $R$ is $N$-1 (https://stacks.math.columbia.edu/tag/0BI1) , then $R^{fn}$ exists and $R^{fn}=\overline R$, which motivates my second question: If $R^{fn}$ exists, then is $\overline R$ module finite over $R$ ? i.e. is $R^{fn}=\overline R$ ?
I'm willing to assume that $R$ has characteristic zero and is local if that helps give any positive answers ...
Finite closure exists if and only if the integral closure is a finite module over $R$. As you observed, if the integral closure is a finite module, it is the finite closure. Conversely, if the finite closure exists, since for any element $x$ in the integral closure, $R[x]$ is a finite module, you see that $x$ must be in the finite closure and thus the integral closure must be contained in the finite closure.