I will prove this theorem.
Theorem: There is no massively palindromic number larger than $7$.
Proof
For $n>105$, it is obvious that $2(\lfloor\sqrt{n}\rfloor-2)^2>2(\sqrt{n}-3)^2=2n-12\sqrt n+18>n$. Therefore, there exists at least $3$ bases, $\lfloor\sqrt{n}\rfloor$, $\lfloor\sqrt{n}\rfloor-1$ and $\lfloor\sqrt{n}\rfloor-2$ which makes $n$ $3$ digit number starting with $1$ in its base. Therefore, $n$ should also end with $1$ in these bases and it follows that $n-1$ is common multiple of $\lfloor\sqrt{n}\rfloor$, $\lfloor\sqrt{n}\rfloor-1$ and $\lfloor\sqrt{n}\rfloor-2$. However, it can be easily proved that the lcm of $3$ consecutive numbers is at least the half of their product. Then one can see that $$(\lfloor\sqrt{n}\rfloor)(\lfloor\sqrt{n}\rfloor-1)(\lfloor\sqrt{n}\rfloor-2)>(\sqrt{n}-1)(\sqrt{n}-2)(\sqrt{n}-3)>n^{3/2}-6n>2(n-1)$$so this is contradiction.
Now, we only need to check the case $n\le 105$. Filtering by base $2$ gives possible candidates$$1, 2, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 73, 85, 93, 99$$
and filtering them by base $3$ gives candidates$$1, 2, 3, 5, 7$$And these numbers are only massively palindromic number. They have only $0$ or $1$ bases which makes them more than $3$ digit number, but they are all palindromes in such bases.
Also, it is known that even palindromic in both bases $2$ and $3$ is extremely rare condition. See this question.
