$\newcommand{\cO}{\mathcal{O}}\newcommand{\fm}{\mathfrak{m}}\newcommand{\fp}{\mathfrak{p}}$ Let $(\cO,\fm)$ be a Noetherian reduced local ring, essentially of finite type over an algebraically closed field $k$. Let $\fp_1,\ldots,\fp_r$ be the minimal prime ideals. Then is it true that $$ \bigcap_{1\leq i\leq r}(\fm^n+\fp_i)=\fm^n $$ for all $n\geq 1$?
Of course we have the inclusion $\bigcap_{1\leq i\leq r}(\fm^n+\fp_i)\supseteq\fm^n$, so we are left to see whether the reverse inclusion holds as well. Here are some considerations: we have a ring map $$ \begin{align*} \varphi\colon\cO&\to\bigoplus_{1\leq i\leq r}\cO/\fp_i\\ f&\mapsto (f+\fp_1,\ldots,f+\fp_r), \end{align*} $$ which is injective as $\bigcap_i\fp_i=0$. Then the question is equivalent to $(\fm^n)^{ec}=\fm^n$, where $\bullet^e$ resp. $\bullet^c$ denotes extension resp. contraction along $\varphi$. Does this hold here?
Note that it holds in baby cases like $\cO=k[x,y]_{(x,y)}/(xy)$. Indeed, in this case the minimal primes are $\fp_1=(x)$ and $\fp_2=(y)$. Suppose that $$ f/g\in ((x,y)^n+(x))\cap ((x,y)^n+(y))=(x,y^n)\cap(x^n,y) $$ for some $f,g\in k[x,y]$ with $g\notin (x,y)$. Then by the above we have in particular $$ f\in (x,y^n)\cap(x^n,y) $$ inside $k[x,y]$. In particular, $f(x,0)\in(x^n)$ and $f(0,y)\in (y^n)$. Therefore, we obtain that $f+(xy)\in (x,y)^n/(xy)$ inside $k[x,y]/(xy)$. After localizing, this implies $f/g\in\fm^n$ inside $\cO$, so we are done.
Counerexample: let $\mathcal{O}=k[x,y]_{(x,y)}/((x-y^2)(x-y^3)).$ The minimal primes are $(x-y^2)$ and $(x-y^3)$. For $n=2,$ $x$ is contained in $\mathfrak{m}^2+(x-y^2)$ and in $\mathfrak{m}^2+(x-y^3)$ but not in $\mathfrak{m}^2$. For, $$\mathcal{O}/\mathfrak{m}^2=[k[x,y]/((x,y)^2+(x-y^2)(x-y^3))]_{(x,y)}=k[x,y]/(x,y)^2$$
and $x\neq 0$ in the right-hand-side ring.