Thanks to Zach Teitler for the comment that this tensor is associated with elementary symmetric polynomial (ESP). I searched for the decomposed form of ESP and found a paper (Power Sum Decompositions of Elementary Symmetric Polynomials by Lee (2015)).

I followed that and all I had to do was to translate the algebraic expressions in the context of tensors. There might be small errors in my part but this is basically what I found.

The $N$-way, $M$-dimensional tensor $\mathcal{Z}$ is equivalent to the ESP $S_{N}^{M}$ -

\begin{align*}
\mathcal{Z}_{p_1 ... p_N}
&= \sum_{\mu (I) = 1}^{R} \: \alpha_{\mu} \: t_{p_1}^{\mu} ... t_{p_N}^{\mu}
\\
S_{N}^{M}
&= \sum_{1 \leq p_1 \leq ... \leq p_N \leq M} X_{p_1} ... X_{p_N}
\quad (2 \leq N \leq M)
\end{align*}

where R is the rank.

The vector $\alpha$ and matrix $t$ are defined as -

\begin{align*}
t_{p}^{\mu (I)}
&= - 1, \quad \mbox{if} \: p \in \{ I \}
\\
&= + 1, \quad \mbox{otherwise}
\end{align*}

\begin{align*}
\alpha_{\mu (I)}
&= (-1)^{n_I} \: \frac{1}{2^{N-1}} \:
\begin{pmatrix}
M - k - n_I - 1 \\
k - n_I
\end{pmatrix}, \quad \mbox{if} \: k = \frac{N-1}{2} \:
\mbox{(odd)}
\\
&= (-1)^{n_I} \: \frac{1}{2^N \: (M-N)} \:
\begin{pmatrix}
M - k - n_I - 1 \\
k - n_I
\end{pmatrix} \: (M - 2 \: n_I), \quad \mbox{if} \: k = \frac{N}{2} \: \mbox{(even)}
\end{align*}
where $ \{ I \} \subset \{ M \} $ and $ n_I = | I | \leq k $. All the subsets are ordered and assigned an integer which belongs to $ \{ \mu (I) \} $. R has an upper bound ($ R \leq \sum_{r=0}^{k} \: \begin{pmatrix}
M \\ r \end{pmatrix}$).