This is not exactly the same thing, but it is similar.

Let $\lambda = (10,10,\dotsc,10)$, and $\mu=(4,4,\dotsc,4)$.
The non-tileability implies that the irreducible character, $\chi^{\lambda}(\mu)$ is $0$. Furthermore, since $\mu$ has all entries equal, turns out that $|\chi^{\lambda}(\mu)|$ is an upper bound on the number of tilings. See the Murnaghan-Nakayama rule for the connection.

Now, $|\chi^{\lambda}(\mu)|$ can in this case be computed via a Hook formula, see On the [Number of Rim Hook Tableaux by Fomin & Lulov][1]

With $n=100$, $r=4$, $m=25$, the first theorem in their paper then states that
$$
f^{\lambda}_r \leq \frac{m! r^m}{(n!)^{1/r}} (f^\lambda)^{1/r} 
$$
and since $|\chi^{\lambda}(\mu)| = f^{\lambda}_r$ in this case,
we get $\approx 2.78073*10^{16}$.

The number $f^{\lambda}_r$ count the number of ways to cover the 
the $10\times 10$-square with rim-hooks of size $4$, and 
*it keeps track of the order one adds the pieces*, such that the first $k$
pieces always form a Young diagram.
This disallow for some configurations, but since the order matters, 
$f^{\lambda}_r$ should be an upper bound.

  [1]: https://link.springer.com/article/10.1007/BF02355806