Of all integers? That's impossible!

Let $f(p) = p^4 - 1 = (p - 1)(p + 1)(p^2 + 1).$ Note that $f(7) = 2^5 \cdot 3 \cdot 5^2$ and $f(11) = 2^4 \cdot 3 \cdot 5 \cdot 61.$ We can show that their GCD, $2^4 \cdot 3 \cdot 5 = 240$ is actually the GCD of all $p^4 -1.$

Since $p$ is odd, $p^2 + 1$ is even. Both $p-1$ and $p+1$ are even, and since they are consecutive even numbers, one is divisible by 4. Thus, $f(p)$ is always divisible by $2^4.$

When divided by 3, $p$ leaves remainder 1 or 2. If 1, then $3|p-1$ and if 2, then $3|p+1$ so $f(p)$ is always divisible by 3. Using the same casework, we can see that $f(p)$ is always divisible by 5.

By Fermat's Little Theorem, $p^4-1 \equiv 0 \mod 5$ and $p^4-1=(p^2)^2 -1 \equiv 0 \mod 3$ . So by the Chinese Remainder Theorem, $p^4-1 \equiv 0 \mod 15$ . Furthermore, $p^4-1 \equiv 0 \mod 16$ for all odd $p$ , however, this is not true modulo $32$ , for if $p \equiv 3 \mod 32$ , then $p^4-1 \equiv 3^4-1 \equiv 80 \equiv 16 \mod 32$ . Therefore, $15 \cdot 16 =240$ must evenly divide the answer. Since $gcd(7^4-1, 11^4-1)=240$ , the answer must be $240$ .

Note that $\gcd(7^4-1,11^4-1)=240$ . Since the Carmichael Lambda of 240 is 4, we know that $a^4\equiv {1}\ \pmod{240}$ for any $a$ that is coprime to 240. Thus the answer is $\boxed{240}$