For every prime $p\geq 5$ one seems to have the congruence
$$(-1)^{(p-1)/2}\prod\_{k=0}^{p-1}{p-1\choose k}\equiv 1-p+\frac{3}{2}p^2-\frac{7}{6}p^3\pmod{p^4}\ .$$
(I have checked all primes up to $5000$.)


The congruence can also be expressed using the easy identities
$$\prod\_{k=0}^n{n\choose k}=\frac{(n!)^{n+1}}{\left(\prod\_{k=0}^nk!\right)^2}
=\frac{(n!)^{n-1}}{\left(\prod\_{k=0}^{n-1}k!\right)^2}\ .
$$


Congruence modulo $p$ is easy. I have however no idea why such a nice congruence holds modulo $p^4$. (I was unable to get something modulo $p^5$: If it exists, it is more complicated since the difference is congruent to $0\pmod{p^5}$ for $p=107$ and $p=433$.)


*It would be nice to have a counterexample or a proof?*



A congruence for a product of binomial coefficients?
At first, $$(-1)^k{p-1\choose k}=\frac{(1-p)(2-p)\cdots (k-p)}{1\cdot 2\cdots k}=\left(1-\frac{p}1\right)\left(1-\frac{p}2\right)\cdots \left(1-\frac{p}k\right)
\\\equiv 1-pe\_1(1,1/2,\ldots,1/k)+p^2 e\_2(1,1/2,\ldots,1/k)-p^3e\_3(1,1/2,\ldots,1/k), \pmod{p^4}$$
where $e\_i$ stands for the $i$-th elementary symmetric polynomial.
Since $(-1)^{(p-1)/2}=\prod\_{k=0}^{p-1}(-1)^k$, this allows to rewrite your product modulo $p^4$ as
$$
A:=\prod\_{k=0}^{p-1}(1-pe\_1(1,1/2,\ldots,1/k)+p^2 e\_2(1,1/2,\ldots,1/k)-p^3e\_3(1,1/2,\ldots,1/k)).
$$
Expand the brackets and use
\begin{align\*}e\_1(1,1/2,\ldots,1/k)&=\sum\_{j\leqslant k} \frac1j,\\
e\_2(1,1/2,\ldots,1/k)&=\sum\_{i<j\leqslant k} \frac1{ij},\\
e\_3(1,1/2,\ldots,1/k)&=\sum\_{\ell<i<j\leqslant k} \frac1{\ell ij}.\end{align\*} We get modulo $p^4$ that
$$
A\equiv 1+pB+p^2C+p^3D,
$$
where $B,C,D$ are some expressions of multi-zeta type. Let's start from $B$:
$$
B=-\sum\_{k=0}^{p-1} \sum\_{j\leqslant k}\frac1j=-\sum\_{j=1}^{p-1}\frac1j\cdot(p-j)=
p-1-p\sum\_{j=1}^{p-1}\frac{1}j.
$$
Since $\sum\_{j=1}^{p-1}\frac{1}j$ is divisible by $p^2$ for $p>3$, the $pB$ term is already known modulo $p^4$: it is congruent to $p^2-p$.


Next, $C$. We have
$$
C=\sum\_{k\_1<k\_2} e\_1(1,1/2,\ldots,1/k\_1)
e\_1(1,1/2,\ldots,1/k\_2)+\sum\_k e\_2(1,1/2,\ldots,1/k)\\
=\frac12\left(\sum\_k e\_1(1,1/2,\ldots,1/k)\right)^2+
\sum\_k \left(e\_2(1,1/2,\ldots,1/k)-\frac12(e\_1(1,1/2,\ldots,1/k))^2\right)\\
=\frac12\left(\sum\_k e\_1(1,1/2,\ldots,1/k)\right)^2-\frac12
\sum\_k e\_1(1,1/2^2,\ldots,1/k^2).
$$
This is of interest modulo $p^2$. We already know
$\sum\_k e\_1(1,1/2,\ldots,1/k)\equiv 1-p\pmod {p^2}$. Analogously we get
$$
\sum\_k e\_1(1,1/2^2,\ldots,1/k^2)=\sum\_j (p-j)/j^2=p\sum\_j 1/j^2-\sum\_j 1/j\equiv 0\pmod {p^2}.
$$
Thus, modulo $p^4$ we get $p^2C\equiv \frac12p^2(1-p)^2\equiv \frac12p^2-p^3$.
Well, we already have your congruence modulo $p^3$, and it remains to prove that $D\equiv -\frac16 \pmod p$. Ok, let's check this.
Denote $a(k)=e\_1(1,1/2,\ldots,1/k)$, $b(k)=e\_2(1,1/2,\ldots,1/k)$,
$c(k)=e\_3(1,1/2,\ldots,1/k)$.
We have
$$
D=-\sum\_{k\_1<k\_2<k\_3} a(k\_1)a(k\_2)a(k\_3)-\sum\_{k\_1\ne k\_2} b(k\_1) a(k\_2)-\sum\_k c(k).
$$
All further congruences are modulo $p$. We already know that $\sum\_k a(k)\equiv 1$. Thus
$$
1=\left(\sum\_k a(k)\right)^3=\sum\_k a(k)^3+6\sum\_{k\_1<k\_2<k\_3} a(k\_1)a(k\_2)a(k\_3)
+3\sum\_{k\_1\ne k\_2} (a(k\_1))^2a(k\_2)\\
=-2\sum\_k a(k)^3+6\sum\_{k\_1<k\_2<k\_3} a(k\_1)a(k\_2)a(k\_3)
+3\left(\sum\_{k\_1} (a(k\_1))^2\right)\left(\sum\_{k\_2}a(k\_2)\right)
$$
and we get
$$
D=-\frac16-\frac13\sum\_k a(k)^3+\frac12\left(\sum\_{k\_1} (a(k\_1))^2\right)\left(\sum\_{k\_2}a(k\_2)\right)-\left(\sum\_{k\_1} b(k\_1)\right)\left(\sum\_{k\_2}a(k\_2)\right)\\+\sum\_k b(k)a(k)-\sum\_k c(k).
$$
At first, we have
$$
\frac12\left(\sum\_{k\_1} (a(k\_1))^2\right)\left(\sum\_{k\_2}a(k\_2)\right)-\left(\sum\_{k\_1} b(k\_1)\right)\left(\sum\_{k\_2}a(k\_2)\right)=
\frac12 \left(\sum\_{k\_2}a(k\_2)\right) \left(\sum\_{k\_1} e\_1(1,1/2^2,\ldots,1/k\_1^2)\right)
$$
which is proved above to be congruent to 0 modulo $p$, and we remain (after multiplication by $-3$) with proving that $p$ divides
$$
\sum\_k a(k)^3-3b(k)a(k)+3c(k)=\sum\_k \sum\_{j\leqslant k} j^{-3}
\equiv \sum\_j j^{-3}\cdot (p-j)
$$
which is indeed 0.


