First, let me point out that having a chain of $3$ such cardinals, with elementary embeddings $$V_{\kappa+1}\to V_{\lambda+1}\to V_{\eta+1},$$ already implies that $\kappa$ is $1$-inaccessible with target $\lambda$ in $V_{\eta+1}$, and so by elementarity $\kappa$ will be $1$-extendible with arbitrarily large targets below $\lambda$, and so $V_\kappa$ must also think that there are unboundedly many $1$-extendible cardinals. So even chains of length three exceed a proper class of $1$-extendible cardinals in consistency strength.

Meanwhile, a $2$-extendible cardinal is sufficient to produce very large chains:

**Theorem.** If $\kappa$ is $2$-extendible, then $V_\kappa$ has
such a proper class chain of $1$-extendibles as you desire.

Proof. Suppose that $\kappa$ is $2$ extendible with target
$\lambda$, so that there is an elementary embedding
$j:V_{\kappa+2}\to V_{\lambda+2}$ with critical point $\kappa$,
for which necessarily $j(\kappa)=\lambda$. It follows that
$V_{\lambda+2}$ can see the embedding $j\upharpoonright
V_{\kappa+1}:V_{\kappa+1}\to V_{\lambda+1}$ and therefore $\kappa$
is $1$-extendible in $V_{\lambda+2}$. It follows that there is a
normal measure one set of $1$-extendible cardinals below $\kappa$.
But furthermore, we can make them cohere in the way you desire in
your question. Namely, since $V_{\lambda+2}$ thinks that $\lambda$
is the target of the $1$-extendibility embedding $j\upharpoonright
V_{\kappa+1}$, and $\lambda=j(\kappa)$, it follows by
elementarity that $V_{\kappa+2}$ thinks that $\kappa$ is the
target of a $1$-extendibility embedding $j_0:V_{\delta+1}\to
V_{\kappa+1}$. Thus, we've achieved a chain of length $2$. By
Zorn's lemma, let us extend this to a maximal set $X\subset\kappa$
(maximal under inclusion) of $1$-extendible-with-target-$\kappa$
cardinals $\delta\lt\kappa$, which is also a chain in your sense.
I claim that $X$ must have size $\kappa$. If not, then $X$ has
size less than $\kappa$. Notice that $V_{\kappa+2}$ is capable of
verifying the property on $X$ we have described, and so by
elementarity, $V_{\lambda+2}$ will think that $j(X)=X$ is a
maximal chain of $1$-extendible-with-target-$\lambda$ cardinals
below $j(\kappa)=\lambda$, which is a chain in your sense. But the
point is that $X\cup\{\kappa\}$ is a strictly larger set of
$1$-extendible-with-target-$\lambda$ cardinals, which is a chain.
This violates the maximality of $X$. Thus, $X$ must be unbounded,
and so we get a proper class chain in $V_\kappa$ of such
cardinals. QED

One can do a similar argument with an alternative weaker hypothesis.

**Theorem.** If $\kappa$ is $2^\kappa$-supercompact, then $V_\kappa$ has a $\kappa$-chain of $1$-extendible cardinals.

Proof. Let $j:V\to M$ be a $2^\kappa$-supercompactness embedding. Thus, $j\upharpoonright V_{\kappa+1}:V_{\kappa+1}\to M_{j(\kappa)+1}$ is a $1$-extendibility embedding inside $M$. So $M$ thinks $\kappa$ is $1$-extendible with target $j(\kappa)$. Let $X\subset\kappa$ be a maximal chain of $1$-extendible-with-target-$\kappa$ cardinals, as in the previous argument. If $X$ is bounded below $\kappa$, then $j(X)=X$, which would contradict maximality, since we may add $\kappa$ to have a strictly larger chain. So $X$ is unbounded, and we've found the desired chain. QED