This is a partial result, which is that there is an $(M,E)$ satisfying all but the end-extension requirement, in place of $M$ being an end-extension of $L_{\gamma_1}$, there is just an $X\subseteq M$ where $(X,E)\cong (L_{\gamma_1},\in)$. Shortly before exercise 3.5.(i), Welch says that it is possible to use Barwise compactness to produce end extensions, but since the $\nu^n$-axioms already form a $\gamma_1$-finite set, it seems difficult in this scenario.
Let $E_n$ be Welch's classes defined in "Weak Systems of Determinacy and Arithmetical Quasi-Inductive Definitions" (2009, definition 15): $E_0$ is the class of $\alpha$ such that there exists $\beta>\alpha$ where $L_\alpha\prec_{\Sigma_2}L_\beta$ (Welch calls these ordinals $\Sigma_2$-extendible, not to be confused with the much larger extendible cardinals), and $E_{n+1}$ is the class of ordinals in $E_0$ which are limit points of $E_n$. Welch mentions that if $\gamma\in E_n$, then there exist ordinals $\mu_n\leq\mu_{n-1}\leq\ldots\leq\mu_0<\nu_0<\ldots<\nu_{n-1}<\nu_n$ such that $\gamma=\mu_n$ and for all $i\leq n$, $L_{\mu_i}\prec_{\Sigma_2}L_{\nu_i}$ (however by well-foundedness of ordinals, being in $\bigcap_{n\in\mathbb N}E_n$ is not enough to give the desired infinite nested intervals.) However, Welch indexes the $\mu$- and $\nu$-ordinals backwards relative to this question, but for finite chains it is equivalent by swapping labels $0$ and $n$, $1$ and $n-1$, etc. So for the rest of this answer I will use the reversed convention $\mu_0\leq\mu_1\leq\ldots\leq\mu_n<\nu_n\ldots<\nu_1<\nu_0$.
Lemma 1: I claim that $L_\alpha\prec_{\Sigma_2}L_\beta$ is not just a $\Pi_2$ property but a $\Sigma_1$ property. The $n=2$ case of theorem 1.8 in Kranakis's "Reflection and partition properties of admissible ordinals" (1980, AML vol 22) is that there is a $\Pi_2$ formula $\varphi_2(\alpha)$ true in $J_\beta$ iff $J_\alpha\prec_{\Sigma_2}J_\beta$, then the formula with free variables $\alpha,\beta$ that is $\varphi_2(\alpha)$ with its quantifiers bounded to $J_\beta$ (i.e. $\exists x(x=J_\beta\land\varphi_2^x(\alpha)$) is $\Sigma_1$. Since all ordinals in this question are $\xi$ such that $\xi=\omega\xi$, $J$ may be replaced with $L$. $\square$
Inductively on $n$, the $\leq$'s relating the ordinals $\mu_k$ may be sharpened to $<$'s as in your question: If $n=0$ it is already sharp. If $n>0$ then we are requiring $\mu_1$ to be $E_{n-1}$, and since $L_{\mu_0}$ satisfies $\forall\xi\exists\sigma(\sigma>\xi\land \sigma\textrm{ is }E_{n-1})$, $L_{\nu_0}$ satisfies this as well. Since $\nu_0$ is a limit point of $E_{n-1}$, choose a new $E_{n-1}$ ordinal in the interval $(\gamma,\nu_0)$ to be the new $\mu_1$.
Lemma 2: First, we inductively show $\gamma_0$ is in $E_n$ for all $n\in\mathbb N$, and that $L_{\gamma_1}\vDash\gamma_0\textrm{ is in }E_n$. For $n=0$: choose a $\gamma<\gamma_1$ such that $L_{\gamma_0}\prec_{\Sigma_2}L_\gamma$ as you mention, then $\gamma_0$ is in $E_0$ and $L_{\gamma_1}$ sees the $\Sigma_2$-elementary-substructurehood. For $n>0$: assuming it has been shown that $L_{\gamma_1}\vDash\gamma_0\textrm{ is in }E_n$, then $L_{\gamma_1}\vDash\textrm{there is an }E_n\textrm{-ordinal}$, and also for any $\xi<\gamma_0$, $L_{\gamma_1}\vDash\textrm{there is an }E_n\textrm{-ordinal}>\xi$. This is a $\Sigma_2$ formula with parameter $\xi\in L_{\gamma_0}$, so by $\Sigma_2$-elementarity, $L_{\gamma_0}$ satisfies this for any $\xi<\gamma_0$ as well. Therefore $\gamma_0$ is a limit point of $E_n$, and $L_{\gamma_1}$ sees this. Since $L_{\gamma_1}$ sees that $\gamma_0$ is also in $E_0$, then it sees that $\gamma_0$ is in $E_{n+1}$. $\square$
In view of the above, given a natural number $n$, let $(\nu^n_k)_{k\leq n}$ be a choice of such ordinals $\nu_k$ from a sequence of $n$ nested intervals initiated by $\gamma_0$. For each $n\in\mathbb N$, $\nu^n$ is a decreasing function with domain $\{0,\ldots,n\}$ (decreasing, since we are using the reversed convention.) Let $\pi$ be a bijection from $\omega$ to $L_{\gamma_1}$, and let $\mathcal L$ be the usual language of set theory extended by countably many constant symbols $c_k$, and a constant symbol $\nu$. Let $\Gamma$ be the set of the following $\mathcal L$-sentences:
- The axioms of KP set theory,
- "$c_j\in c_k$" whenever $\pi(j)\in\pi(k)$, and "$c_j\notin c_k$" whenever $\pi(j)\notin\pi(k)$,
- For each $n\in\mathbb N\setminus\{0\}$, the "$\nu^n$-axiom": the sentence "$\nu$ is a decreasing function to a set of ordinals, $\textrm{ran}(\nu)$ is an ordinal $\geq n+1$, and there is an increasing sequence of ordinals $(\mu_k)_{k\leq n}$ such that for all $k\leq n$, $L_{\mu_k}\prec_{\Sigma_2}L_{\nu_k}$". In these axioms $\nu_k$ stands for $\nu(k)$, i.e. "$\nu$ applied to $k$".
In order to show any finite subset of $\Gamma$ is satisfiable, let $n$ be largest such that the $\nu^n$-axiom is present in the finite subset. $\nu^n$ is in $L_{\gamma_1}$, since admissible sets are closed under finite function formation via pairing and union. Since $L_{\gamma_1}\vDash``\gamma_0\textrm{ is in }E_n"$ (lemma 2), $L_{\gamma_1}$ sees that the set $\nu^n\in L_{\gamma_1}$ satisfies the $\nu^n$-axiom, so $(L_{\gamma_1},\in,(\pi(k))_{k\in\mathbb N},\nu^n)$ satisfies the finite subset of $\Gamma$. In particular, if choosing in $L_{\gamma_1}$ witnesses $(\mu_k)_{k\leq n}$ mentioned in the $\nu^n$-axiom, a possible choice for $\mu_0$ is $\gamma_0$.
To produce an ill-founded $(\widetilde M,E)$ with an infinite chain, apply the compactness theorem to get a model $(\widetilde M,E,(c_k)_{k\in\mathbb N},f)$. $(\widetilde M,E)\vDash KP$, and the first set of $c$-axioms requires that there is a part of $(\widetilde M,E)$ (i.e. $\{c_k\mid k\in\mathbb N\}$) realizing the graph of $\in$ restricted to $L_{\gamma_1}$. $d$ is in $\widetilde M$, and since every $\nu^n$-axiom holds, $f$ is an infinite decreasing sequence of "$(\widetilde M,E)$-ordinals". Any choice of $\mu_0$ will be a $\widetilde M$-ordinal initiating a tower of infinite nested $\Sigma_2$-extendibility intervals.
To produce such a model $M$ with $\gamma_1\notin M$, take a model isomorphic to $\widetilde M$ such that $\gamma_1\notin M$ and $M\cap L_{\gamma_1}=L_{\gamma_1}$.
