How much bigger is 3↑↑↑↑3 compared to 3↑↑↑3? 3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?
By definition, $3\uparrow\uparrow\uparrow\uparrow 3 = 3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow 3))$.
\begin{matrix}
   3\uparrow\uparrow\uparrow 3= &amp; \underbrace{3^{3^{3^{3^{\cdot^{\cdot^{\cdot^{\cdot^{3}}}}}}}}} \\
   &amp; \mbox{7,625,597,484,987 copies of 3}
  \end{matrix}

which should make it a little clearer how much bigger $3\uparrow\uparrow\uparrow\uparrow 3$ is than $3\uparrow\uparrow\uparrow 3$. It's hard to even give a more concrete answer because the numbers of powers of $3$ get unreasonably large to compute or type.
3^^^3 is equal to 3^^3^^3 which is equal to 3^^(3^3^3) 
3^^(3^3^3)= 3^^(3^27)= 3^^(7.6 trillion)= 3^3^3^3... 7.6 trillion times.
Let's represent this mind boggling number as x.

3^^^^3 is equal to 3^^^3^^^3 which is equal to 3^^^(x)
which is equal to 3^^3^^3^^3...(with x ^^3's)
Two threes: 3^^3 is around 7.6 trillion
Three threes: 3^^3^^3= x
Four threes: 3^^3^^3^^3= 3^3^3...(with x ^3's)
At five threes I can't even list out the exponentiation.

You get how if there were, say, a googleplex threes all tetrated, it would be completely absurd.

But we know that x is far far far (and I can not stress this enough) far far larger than a googleplex, or even a number with a googleplex digits. 

In conclusion:
3^^^^3= 3^^^3^^^3
=3^^3^^3^^3...(with x 3's)
And because at even just 3^^3^^3^^3^^3 I can't list out the exponentiation, it is impossible to do so with x 3's.

So yes, 3^^^^3 is so large it is simply around just 3^^^^3 times larger than 3^^^3.

(although it really is just 3^^^3^^^3.)