First I will rephrase the question, and then I will solve the rephrased problem.
Let ${\rm Part}_n$ denote the set of set partitions of $[n]=\{1,2,\ldots,n\}$.
For $\Theta,\Pi\in {\rm Part}_n$, we will say that $\Theta$ is finer than $\Pi$ and write $\Theta\le \Pi$, when
$$
\forall A\in\Theta, \exists B\in\Pi, A\subset B\ .
$$
The $\subset$ notation here allows equality.
This defines the well-known partition lattice $({\rm Part}_n,\le)$. The smallest element denoted by $\hat{0}$ is the partition made of singletons. The largest element denoted by $\hat{1}$ is the single block partition. We will use $\vee$ to denote the join or supremum of two partitions in ${\rm Part}_n$.
The lattice has a Möbius function $\mu(\Theta,\Pi)$ given by the entries of the matrix
$(\mu(\Theta,\Pi))_{\Theta,\Pi\in{\rm Part}_n}$ which by definition is the inverse of the matrix $(\mathbf{1}\{\Theta\le \Pi\})_{\Theta,\Pi\in{\rm Part}_n}$.
We used the notation $\mathbf{1}\{\cdots\}$ for the indicator function equal to $1$ if the enclosed condition is true, and equal to $0$ if the condition is false.
An explicit formula for the Möbius function is given by
$$
\mu(\Theta,\Pi)=\prod_{A\in\Pi}\left[(-1)^{s(\Theta,A)-1}(s(\Theta,A)-1)!\right]
$$
where $s(\Theta,A)$ is the number of blocks of $\Theta$ contained in $A$.
The following lemma reflects the graded/ranked property of the partition lattice.
Lemma:
If $\Theta_1,\Theta_2\in{\rm Part}_n$, then $|\Theta_1\vee\Theta_2|\ge|\Theta_1|+|\Theta_2|-n$.
Proof: For $i=1,2$, pick $F_i$ a forest (subgraph of the complete graph with no circuit) whose partition of connected components is $\Theta_i$. We have $|\Theta_i|=n-|F_i|$.
By eliminating redundant edges which create circuits instead of new connections, pick a forest $F_3\subset F_1\cup F_2$ such that the partition of connected components of $F_3$ is exactly $\Theta_1\vee\Theta_2$. Since $|F_3|\le |F_1|+|F_2|$, the wanted inequality follows.
The notation $(x)_k:=x(x-1)\cdots(x-k+1)$ is for the falling factorial.
We now define
$$
K_{\Pi}(N):=\sum_{\Theta\ge \Pi}(-1)^{|\Theta|}(|\Theta|-1)!
\times\prod_{A\in\Theta}\left[\prod_{j=1}^{|A|-1}\left(1-\frac{j}{N}\right)\right]
$$
$$
=\sum_{\Theta\ge \Pi}(-1)^{|\Theta|}(|\Theta|-1)!
\times\prod_{A\in\Theta}\left(\frac{(N-1)_{|A|-1}}{N^{|A|-1}}\right)
$$
$$
=\sum_{\Theta\ge \Pi}(-1)^{|\Theta|}(|\Theta|-1)!
\times\prod_{A\in\Theta}\left(\frac{(N)_{|A|}}{N^{|A|}}\right)
$$
$$
=-N^{-n}\ P_{\Pi}(N)
$$
where we introduced the polynomial, in $N$,
$$
P_{\Pi}(N):=\sum_{\Theta\ge \Pi}\mu(\Theta,\hat{1}) \prod_{A\in\Theta} (N)_{|A|}\ .
$$
If I understood correctly, the question is to show the existence of a constant $C_n$, only depending on $n$, such that for all $N>0$
$$
|K_{\Pi}(N)|\le C_n N^{1-|\Pi|}\ ,
$$
i.e.,
$$
|P_{\Pi}(N)|\le C_n N^{n-|\Pi|+1}\ .
$$
The means the degree of the polynomial $P_{\Pi}$ is at most $n-|\Pi|+1$.
Using Stirling numbers of the first kind, and their definition in terms of permutations in the symmetric group $\mathfrak{S}_k$, we have
$$
(x)_k=\sum_{\rho\in\mathfrak{S}_k}{\rm sgn}(\rho) x^{c(\rho)}
$$
where $c(\rho)$ is the number of cycles of the permutation $\rho$.
Grouping permutations according to the set partition corresponding to their cycles we have
$$
(x)_k=\sum_{\Theta\in{\rm Part}_k}(-1)^{k-|\Theta|}x^{|\Theta|}
\prod_{A\in\Theta}(|A|-1)!
$$
$$
=\sum_{\Theta\in{\rm Part}_k}\mu(\hat{0},\Theta)x^{|\Theta|}
$$
where the Möbius function is that of ${\rm Part}_k$.
By taking the product of the previous identity over the blocks of some partition $\Theta\in{\rm Part}_n$, and expanding, we have
$$
\prod_{A\in\Theta} (N)_{|A|}=\sum_{\Gamma\le\Theta}\mu(\hat{0},\Gamma)N^{|\Gamma|}
$$
where the Möbius function is that of ${\rm Part}_n$.
As a result,
$$
P_{\Pi}(N)=\sum_{\Theta\ge\Pi}\mu(\Theta,\hat{1})
\left(\sum_{\Gamma\le \Theta}\mu(\hat{0},\Gamma)N^{|\Gamma|}\right)
$$
$$
=\sum_{\Gamma\in{\rm Part}_n}\mu(\hat{0},\Gamma)N^{|\Gamma|}
\left(\sum_{\Theta\ge \Pi\vee\Gamma}\mu(\Theta,\hat{1})\right)
$$
$$
=\sum_{\Gamma\in{\rm Part}_n}\mu(\hat{0},\Gamma)N^{|\Gamma|}\mathbf{1}\{\Pi\vee\Gamma=\hat{1}\}
$$
and the degree bound immediately follows from the Lemma.
As an aside, here are two more ways to write the polynomial $P_{\Pi}(N)$.
$$
P_{\Pi}(N)=\sum_{\rho\in\mathfrak{S}_n}\mathbf{1}\{\langle\tau,\rho\rangle\ {\rm transitive}\} {\rm sgn}(\rho)N^{c(\rho)}\ .
$$
Here $\tau$ is a fixed permutation whose cycle partition is $\Pi$. The sum is over permutations $\rho$ such that the subgroup of the symmetric group, generated by $\tau$ and $\rho$, acts transitively on $[n]$.
$$
P_{\Pi}(N)=\sum_G \mathbf{1}\{G\ \Pi-{\rm connects}\ [n]\}
(-1)^{|G|}N^{C(G)}
$$
where the sum is over subgraphs $G$ of the complete graph $K_n$, seen as a set of edges, with the condition that $G$ together with a forest that has $\Pi$ as partition of connected components, connects the entire set $[n]$. Here $C(G)$ is the number of connected components of $G$. The last formula is related to the Whitney-Tutte-Fortuin-Kasteleyn representation of the chromatic polynomial (for the complete graph). See this article by Sokal, in particular for the relation to rigorous statistical mechanics.
Addendum: One has the more precise result that the degree of $P_{\Pi}(N)$ is exactly $n+1-|\Pi|$, because the sign of $\mu(\hat{0},\Gamma)$ is $(-1)^{n-|\Gamma|}$. The coefficient of $N^{n+1-|\Pi|}$ is
$$
(-1)^{|\Pi|-1}\sum_{\Gamma\in{\rm Part}_n}
\mathbf{1}\{|\Gamma|=n+1-|\Pi|,\Pi\vee\Gamma=\hat{1}\}
\prod_{A\in\Gamma}(|A|-1)!
$$
which is easily seen to be nonzero. I don't know if one can actually compute its value.
