There is no smallest infinitely large prime

Hint: Suppose toward a contradiction that $p$ is the smallest infinite prime. There is a prime between zero and $p$. Moreover, if there is a prime between $n$ and $p$, then there is a prime between $n+1$ and $p$. Therefore...

Hint: Use the fact that if $x\gt 1$, there is always a prime between $x$ and $2x$. (You will want to show that for any infinite $y$ there is an infinite $w$ such that $2w\le y$.)

Remarks: The "fact" is usually called Bertrand's Postulate, and has been a theorem since about $1850$.

Every recursive function is definable, so one could travel through the $n!$ route you suggested.

I think appealing to overspill or induction in particular is overkill here; being an elementary extension is all we need.

First prove that a $\mathfrak B$-natural $n$ that is not infinitely large is in fact a standard element of $\mathbb N$. By assumption there is a standard real $r$ such that $n<r$, and then we also have $ n < m $ where $m$ is some standard natural larger than $r$. But the fact that every natural less than $m$ is one of $0,1,2,\ldots,m-1$ is a finite first-order sentence, and since it is true in $\mathfrak R$ it is also true in the elementary extension $\mathfrak B$. So $n$ is one of those particular standard naturals.

Next, in $\mathfrak R$ it is true that for every prime $p$ that doesn't equal $1+1$, there is a largest prime less than $p$. This is a first-order property and therefore true in $\mathfrak B$ too.

Therefore, if $p$ is any infinitely large prime, it has an immediate predecessor prime $q$.

All we now need to know is that $q$ itself is infinitely large. However, if $q$ is not infinitely large, then it's standard. And then since it is true in $\mathfrak R$ that the only prime that has $q$ as its immediate predecessor is such-and-such, this is also true in $\mathfrak B$, and $p$ must equal such-and-such. Which contradicts the assumption that $p$ was infinitely large.