I believe this is a variation of related questions. Define a falling exponent of a factorial $n!$ to be $1^{n!} \times 2^{n!-1} \times 3^{n!-2} \times \dots \times n!^1$. Thus for $n=3$ you get $1^6 * 2^5 * 3^4 * 5^2 * 6^1=24883200$.
One could solve by doing all the multiplication.
One could solve by adding the logs.
One could find an estimation similar to Stirling's for factorials.
Do you think an accurate estimation formula can be found?
$$ \begin{align} 1^{n!} \times 2^{n!-1} \times \dots n!^{n!-(n!-1)} &= \frac{1^{n!}\times2^{n!}\dots n!^{n!}}{1\times 2^1 \times 3^2 \dots \times n!^{n!-1}} \\ &=(n!)!^{n!}\frac{1}{(n!)!\times\frac{(n!)!}{2!}.\frac{(n!)!}{3!}\dots\frac{(n!)!}{(n!-1)!}} \\ &=(n!)!^{n!}\frac{1!\times2!\times\dots\times(n!-1)!}{(n!)!^{n!-1}} \\ &=1!\times2!\times\dots\times(n!-1)!\times (n!)! \end{align} $$
For $n = 3$, it is $1! \times 2! \dots \times 6! = 2 \times 6 \times 24 \times 120 \times 720 = 24883200$
The formula is $\prod_{i=1}^{n!} i!$. This function is closely related to the superfactorial.
WolframAlpha says that $\prod_{i=1}^{n!}i!=G(x!+2)$, where $G$ is the Barnes-G-function. We can use the asymptotic expansion of $G$ from its Wikipedia article with $z=n!+1$ to find an approximation for your product. The expansion goes on even further so you could make it more accurate if you wanted.
$$\small\prod_{i=1}^{n!}i!\sim\exp\left[\frac{1}{12}-\ln(A)+\frac{n!+1}{2}\ln(2\pi)+\left(\frac{(n!+1)^2}{2}-\frac{1}{12}\right)\ln(n!+1)-\frac{3}{4}(n!+1)^2\right]\quad *$$
I've shown its accuracy for a few values of $n$ in the following table.
$$\begin{array}{|c|c|c|c|} \hline n & \text{Exact} & \text{Approximate} & \text{Relative Error}\\ \hline 3 & 2.488\times10^7 & 2.489\times10^7 & 4.02\times10^{-4} \\ \hline 4 & 1.174\times10^{243} & 1.174\times10^{243} & \approx0 \\ \hline 5 & 1.569\times10^{10\,526} & 1.569\times10^{10\,526} & \approx0 \\ \hline \end{array}$$
So there's a great approximation to your product.
* Where $A=1.28\ldots$ is the Glaisher–Kinkelin constant.
I also tried to make an approximation by adapting Stirling's approximation but it didn't go perfectly. I'll share it here for anyone else to use, if they can,
$$\begin{align} p=\mathrm{sf}(n!)&=\prod_{i=1}^{n!}i! \\ \ln(p)&=\sum_{i=1}^{n!}\ln(i!) \\ \sum_{i=1}^{n!}\left[i\ln\left(\frac{i}{e}\right)+1\right]\leq\ln(p)&\leq\sum_{i=1}^{n!}\left[(i+1)\ln\left(\frac{i+1}{e}\right)+1\right]\quad** \\ e^{n!+\sum_{i=1}^{n!}i\ln\left(\frac{i}{e}\right)}\leq p&\leq e^{n!+\sum_{i=1}^{n!}(i+1)\ln\left(\frac{i+1}{e}\right)} \\ e^{n!}\prod_{i=1}^{n!}e^{i\ln\left(\frac{i}{e}\right)}\leq p&\leq e^{n!}\prod_{i=1}^{n!}e^{(i+1)\ln\left(\frac{i+1}{e}\right)} \\ e^{n!}\prod_{i=1}^{n!}\left(\frac{i}{e}\right)^i\leq p&\leq e^{n!}\prod_{i=1}^{n!}\left(\frac{i+1}{e}\right)^{i+1} \\ \frac{e^{n!}}{e^1\cdot e^2\cdot\ldots \cdot e^{n!}}\prod_{i=1}^{n!}i^i\leq p&\leq \frac{e^{n!}}{e^2\cdot e^3\cdot\ldots \cdot e^{n!+1}}\prod_{i=1}^{n!}(i+1)^{i+1} \\ e^{n!-\frac{n!(n!+1)}{2}}H(n!)\leq p&\leq e^{n!-\frac{n!(n!+1)}{2}-1}H(n!+1)\quad*** \end{align}$$
We can take the geometric mean of these bounds to get an approximation for $p$. I tried it with an arithmetic mean but the bounds are so far apart that the arithmetic mean has roughly the same exponent as the upper bound,
$$ \begin{align} p&\approx H(n!)\left(e^{\frac{n!(1-n!)}{2}}e^{\frac{n!(1-n!)-1}{2}}(n!+1)^{n!+1}\right)^{1/2} \\&\approx H(n!)(n!+1)^\frac{n!+1}{2}\left(e^{n!(1-n!)-1/2}\right)^{1/2} \\&\approx H(n!)(n!+1)^\frac{n!+1}{2}e^\frac{2n!(1-n!)-1}{4} \end{align} $$
However, these steps could equally have been applied to the superfactorial, and replacing the superfactorial with the hyperfactorial isn't much better, so in a sense this work is redundant, though we could now try asymptotic approximations to $H(n)$. It's difficult to see how accurate this approximation is graphically, since neither curve fits easily onto a graph. Regardless, for $n=3,4,5$, the exact value of $p$ and its approximation are shown in the following table:
$$\begin{array}{|c|c|c|c|} \hline n & \text{Exact} & \text{Approximate} & \text{Relative Error} & \text{Calculation} \\ \hline 3 & 2.488\times10^7 & 8.715\times10^8 & 34.0 & \text{http://bit.ly/2cwYgcB} \\ \hline 4 & 1.174\times10^{243} & 1.772\times10^{249} & 1.51\times10^6 &\text{http://bit.ly/2cIND7Y} \\ \hline 5 & 1.569\times10^{10\,526} & 5.180\times10^{10\,556} & 3.30\times10^{30} &\text{http://bit.ly/2cIwjl9}\\ \hline \end{array}$$
Is it a good approximation? No, in fact the relative error is growing so it seems to be diverging. But it begins reasonably well and it's a start. Maybe someone else can adapt it to make it work.
** By the derivation of Stirling's approximation.
*** Where $H(x)$ is the Hyperfactorial function.
