The drinking principle, namely $$ \exists x (p(x) \rightarrow \forall y p(y)) ,$$ aka the drinker paradox, can be infered using the rules of classical logic.
Is it possible to intuitionistically infer a formula which is not intuitionistically inferrable (such as $\lnot F \lor \lnot\lnot F$ or $\lnot F \lor F$ or $\lnot\lnot F \rightarrow F$ or $\lnot \exists x \lnot F \rightarrow \forall x F$ ...) from the drinking principle?
If not, how do you prove it can't be inferred with the intuitionistic rules?
The drinker principle is not so strong that assuming it as an axiom will in itself yield classical logic.
One argument for this is that if $\varphi$ does not contain $x$ free, then $\varphi \to (\forall x)\varphi$ is intuitionistically valid. In particular, the drinker principle itself is intuitionistically valid in the special case where $p(x)$ does not mention $x$.
Thus, suppose that intuitionistic first-order logic plus Drinker proves $p\lor \neg p$ where $p$ is a nullary predicate letter. Take the supposed proof and replace every atomic formula other than $p$ by $\top$ everywhere in the proof. This replacement both preserves all inference rules, and all axioms are still provable after the replacement (equality axioms become things like $\top\to\top$ or $\top\to\top\to\top$, which need short proofs of themselves now). Since there are no terms anywhere anymore, every instance of the Drinker axiom becomes $(\exists x)(\psi\to(\forall x)\psi)$ where $\psi$ does not contain $x$, which is provable without the Drinker axiom. Thus, we can construct a pure intuitionistic proof of $p\lor\neg p$ (which was not affected by the replacement) -- and this is known to be impossible.
To show that Drinker is not itself intuitionisticaly valid, we can show a Kripke model where it is false.
The worlds are $w_0$, $w_1$, $w_2$ with $w_0<w_1$ and $w_0<w_2$, the domain is $\{1,2\}$ everywhere, and $$ w_1\Vdash p(1) \qquad w_2\Vdash p(2) $$ Then $(\forall x)p(x)$ is false everywhere. So $p(1)\to(\forall x)p(x)$ is false in $w_1$ and $p(2)\to(\forall x)p(x)$ is false in $w_2$, and both of these are false in $w_0$. So there cannot be any witness of $p(x)\to(\forall x)p(x)$ in $w_0$, so the Drinker applied to $p$ is false in $w_0$.