Note: the answer here stems from the comment of user fedja above.

The following Lemma is another version of the Riesz–Thorin interpolation theorem (See Lemma 8.5 in this book).

**Lemma.** Let $(X_i , \mathfrak M_i , μ_i )$, $i = 0,1, 2, . . . , n$ be measure spaces.
Let $V_i$ represent the complex vector space of simple functions on $X_i$. Suppose that
$$\Lambda : V_1\times V_2 \times \cdots \times V_n \to V_0$$
is a multilinear operator of types $p_0$ and $p_1$ where $p_0,p_1 \in [1,\infty]$, with constants $C_0$ and $C_1$, respectively. i.e.,
$$
\|\Lambda(f_1 , f_2 , . . . , f_n) \|_{p_i} \leq C_i
\|f_1\|_{p_i}
\|f_2 \|_{p_i}
\cdots \|f_n\|_{p_i} \tag{1}
$$
for $i=0,1$. Let $t \in [0, 1]$ and define
$$\frac 1{p_t} := \frac{1-t}{p_0}+\frac t{p_1}$$
Then, $\Lambda$ is of type $p_t$ with constant $C_t :=C_0^{t-1}C_1^t$, that is,
(1) holds true for $i=t$.

Now, Let $(X_i , \mathfrak M_i , μ_i )$ be a uniform measure on $[n] := \{1,\ldots,n\}$ for $i=1,2$, and be a uniform measure on $[n^2]$ for $i=0$. In this case, we have $V_1= V_2=\mathbb{C}^n$, and $V_0 = \mathbb{C}^{n^2}$. Define $\Lambda : V_1\times V_2 \to V_0$, by $[\Lambda(a,b)]_{i,j} := a_ib_j -a_jb_i$ for $a,b\in \mathbb{C}^n$ and $1\leq i,j\leq n$. Note that, in this setting, we have

$$
\|\Lambda(a,b)\|_p^p = \frac 2{n^2} \sum_{1\leq i<j\leq n} |a_i b_j - a_j a_i|^p\\
\|a\|_p^p = \frac 1n \sum_i |a_i|_p, \quad \|b\|_p^p = \frac 1n \sum_i |b_i|_p
$$

Next, $\Lambda$ is of type $p_0=0$ and $p_1=1$, with constants $C_0=C_1=2$. Above Lemma implies that $\Lambda$ is of type $p$, for every $1<p<2$, with constant $2$.

Also $\Lambda$ is of type $p_0=2$,and $p_1=\infty$, with constant $C_0=2$ and $C_1=4$, respectively. Above Lemma implies that $\Lambda$ is of type $p$, for every $2<p<\infty$, with constant $2 \times 2^{(p-2)/p}$.