okay, this is silly, but I can't for the life of me figure out what's wrong with the following "proof":
Claim: if $B$ is an $(n-1)$-type, then $(n\text{-conn}(A) \to B) \simeq (\text{isCntr}(A) \to B)$.
"proof": We have
\begin{align*}(n\text{-conn}(A) \to B) & \equiv ((\sum_{x : \vert A\vert_n} \prod_{y : \vert A\vert_n} x = y) \to B)\newline
& \simeq (\prod_{x : \vert A\vert_n} \prod_{y : \vert A \vert _n} x = y \to B) \end{align*}
by the universal property of $\sum$ types. Now, since $B$ is an (n-1)-type, it in an n-type, and so is every function type into $B$. So we may apply the universal property of $\vert A\vert_n$ to obtain:
\begin{align*}(\prod_{x : \vert A\vert_n} \prod_{y : \vert A \vert_n} x = y \to B) & \simeq(\prod_{x : A} \prod_{y : A} x = y \to B) \newline
&\simeq ((\sum_{x : A} \prod_{y : A} x = y) \to B) \newline
&\equiv (\text{isCntr}(A) \to B)\end{align*}
This concludes the "proof".
However, this cannot be the case, since this allows us to prove that $! : A \to 1$ is an equivalence for any $n>-1$-connected type $A$. Since $\text{iseqv}(!)$ is a $-1$-Type, we have $(0\text{-conn}(A)\to \text{iseqv}(!))\simeq (\text{isCntr}(A) \to \text{iseqv}(!))$. However, this latter type is always inhabited.
Best Answer
Your problem is that you placed parentheses incorrectly.
The correct placement of parentheses for the critical equivalence is
$$\left(\left(\sum\limits_{x : C} \prod\limits_{y : C} x = y\right) \to B\right) \simeq \left(\prod\limits_{x : C} \left(\left(\prod\limits_{y : C} x = y\right) \to B\right)\right)$$
You apply this identity to both $A$ and $|A|_n$, but you do so with incorrect placement of parentheses. With the correct placement, it is clear that your argument fails.