Why the antisymmetrizer operator works
The following explains why an antisymmetrizer works to create an antisymmetric tensor from an arbitrary tensor, going into a lot of detail and trying to skip very few steps.
The antisymmetrizer operator for a tensor of rank $n$ is:
\[\mathcal{A} \equiv \frac{1}{n!} \sum_{\pi \in S_n} \operatorname{sgn}(\pi) \pi.\]For example, for a tensor $\omega$ of rank $5$ with the indexes $a_{1..5}$:
\[\begin{align} \mathcal{A} \omega_{a_1a_2a_3a_4a_5} = \frac{1}{5!}\sum_{\pi \in S_5} \operatorname{sgn}(\pi)\omega_{a_{\pi(1)},a_{\pi(2)},a_{\pi(3)},a_{\pi(4)},a_{\pi(5)}}. \tag{1}\label{eq1} \end{align}\]WLOG, let’s apply the permutation $(12)$ to $A \omega$ using an example with rank $5$. Since $(12)$ is a linear operator, we have
\[\begin{align} (12) \mathcal{A} \omega_{a_1a_2a_3a_4a_5} &= \frac{1}{5!}\sum_{\pi \in S_5} \operatorname{sgn}(\pi) (12) \omega_{a_{\pi(1)},a_{\pi(2)},a_{\pi(3)},a_{\pi(4)},a_{\pi(5)}} \\ &= \frac{1}{5!}\sum_{\pi \in S_5} \operatorname{sgn}(\pi) \omega_{a_{\pi(2)},a_{\pi(1)},a_{\pi(3)},a_{\pi(4)},a_{\pi(5)}}.\tag{2}\label{eq2} \end{align}\]The effect of $(12)$ was to permute the first and second slots of $\omega$. Note that $(12)$ acted directly on the tensor $\omega$, and did not act in composition with the permutation $\pi$. If there were a composition of two permutations (e.g., something like $\pi (12)$), then the order of the operations matters because permutations are in general not commutative. But in this specific equation, we don’t have that issue.
Consider a specific permutation $\pi_0$ and define $\phi = \pi_0 (1 2)$, so that $\operatorname{sgn}(\pi_0) = -\operatorname{sgn}(\phi)$.
\[\begin{aligned} \omega_{a_{\phi(2)},a_{\phi(1)},a_{\phi(3)},a_{\phi(4)},a_{\phi(5)}} &= \omega_{a_{\pi_0(1)},a_{\pi_0(2)},a_{\pi_0(3)},a_{\pi_0(4)},a_{\pi_0(5)}} \\ -\operatorname{sgn}(\phi) \omega_{a_{\phi(2)},a_{\phi(1)},a_{\phi(3)},a_{\phi(4)},a_{\phi(5)}} &= \operatorname{sgn}(\pi_0) \omega_{a_{\pi_0(1)},a_{\pi_0(2)},a_{\pi_0(3)},a_{\pi_0(4)},a_{\pi_0(5)}} \\ \end{aligned}\]The right side of the last line corresponds to a term in Eq. $\ref{eq1}$, and the left side of the last line corresponds to $-1$ times a term in Eq. $\ref{eq2}$. (Since the $\pi$ permutations in Eqs. $\ref{eq1}$ and $\ref{eq2}$ are dummy variables in sums, they can be renamed to $\pi_0$ or $\phi$.) So there is a one to one correspondence between each term in Eq. $\ref{eq1}$ and the negative of each term in Eq. $\ref{eq2}$, which means that Eq. $\ref{eq1}$ is the negative of Eq. $\ref{eq2}$. This shows that $A\omega$’s sign is flipped via a $(12)$ permutation.
Therefore applying any odd permutation to $A\omega$ flips its sign. Therefore the antisymmetrizer operator indeed yields an antisymmetric tensor when applied to an arbitrary tensor.