Edit (7 May, 2014): With the help of Google Translate, I made a rough translation of Ressayre's statements quoted above:
Please feel free to edit the text for any improvements on the translation.
Edit (9 May, 2014): Let me describe the problems I faced in more details below.
Start with a nonstandard $\omega$-model $M\models\rm ZFC+V{=}L$. Suppose $\varphi(v,x)$ is a $\Pi_n$-formula for which $$A=\{x\in\mathrm L_I^M:\mathrm L_I^M\models\exists v\ \varphi(v,x)\}\not=\varnothing.$$
I can go below $I$ and look at some $\mathrm L_\beta^M\prec_{\Pi_n}\mathrm L_I^M$ for which $A\cap\mathrm L_\beta^M\not=\varnothing$. (I am not sure whether I can have more elementarity while keeping the non-existence of long $\Pi_n$-elementary chains.) We are done if $A\cap\mathrm L_\beta^M$ turns out to be definable in $M$, because then we can apply foundation to it. One way to make this set definable is to find $b\in M$ such that $$\forall x\in\mathrm L_\beta^M\ \bigl( \exists v\in\mathrm L_I^M\ M\models\varphi(v,x) \Leftrightarrow M\models\exists v\in b\ \varphi(v,x) \bigr).$$ As $I$, and hence $\mathrm L_I^M$, may not be definable in $M$, it is not clear how this can be achieved. Actually, it seems possible that $$\forall\delta\in I\quad \exists x\in A\cap\mathrm L_\beta^M\quad \forall v\in\mathrm L_\delta^M\quad M\models\neg\varphi(v,x).$$
Perhaps we should consider $A_\beta=\{x\in\mathrm L_\beta^M:\mathrm L_\beta^M\models\exists v\ \varphi(v,x)\}$, where $\mathrm L_\beta^M\prec_{\Pi_n}\mathrm L_I^M$. This is definable in $M$, and so if it is nonempty, then it has an $\in$-minimal element. However, this $\in$-minimal element may not be an $\in$-minimal element of $A$ because of the possibility that $A_\beta\subsetneq A\cap\mathrm L_\beta^M$.
Well, let us go above $I$ and look at $\mathrm L_\varepsilon^M\succ_{\Pi_n}\mathrm L_I^M$. (Again, I am not sure whether I can have more elementarity while keeping the non-existence of long $\Pi_n$-elementary chains.) Then $A_\varepsilon$, defined in the same way as $A_\beta$ in the previous bullet point, has an $\in$-minimal element. Suppose, out of goodwill, that we can find one such $\in$-minimal element $a\in\mathrm L_I^M$. This would satisfy the desired condition that $x\not\in A$ for all $x\in a$. However, we do not know whether $a\in A$ because we do not have enough elementarity between $\mathrm L_\varepsilon^M$ and $\mathrm L_I^M$.
Alright, perhaps we should also bound the witnesses for $A$: every $\mathrm L_\varepsilon^M\succ_{\Pi_n}\mathrm L_I^M$ satisfies $$\exists x,v\ \bigl( \varphi(v,x)\wedge\forall x',v'(\varphi(v',x')\rightarrow x'\not\in x \bigr).$$ Assuming $I$ is not definable in $M$, this underspills to give $\mathrm L_\delta^M\prec_{\Pi_n}\mathrm L_I^M$ satisfying the displayed sentence above. This does not provide what we want because it only tells us $A_\delta$ has an $\in$-minimal element (which we already saw is apparently not sufficient).
A similar argument shows it is sufficient to prove the existence of some $\mathrm L_\varepsilon^M\succ_{\Pi_n}\mathrm L_I^M$ such that for every $\mathrm L_{\varepsilon'}^M\prec_{\Pi_n}\mathrm L_\varepsilon^M$ above $\mathrm L_I^M$, the set $A_{\varepsilon'}$ contains an $\in$-minimal element of $A_\varepsilon$. This statement is not apparent because even though we can find $a\in\mathrm L_{\varepsilon'}^M$ that is $\in$-minimal for $A_\varepsilon$, we cannot guarantee this $a$ to be an element of $A_{\varepsilon'}$.
…
I could go on, but perhaps this is already a little too much.