3^(2^85) > 2^(2>85), and that latter has 2^85 bits. Any computation attempting to deal with a number that big is going to need substantially more atoms than are available on this planet. So yes, of course this will overflow.

The point of the formula is to give something correct to more places than you can compute. To understand it rewrite 9^(-(4^(67))) as (3^2)^(-(4^(67))) = 3^(-(24^(67))). This is the same as 3^(-(2^85)). So you have an expression of the form

(1+1/n)^n

where n=3^(-(2^85)). To estimate the error between this and E, take logs and see how close we get to 1, then use that to get the multiplicative factor by which we might be off. So the log is of the form n*Log[1+1/n] with n again as above. Expand as a series, truncate at second order, this is n*(1/n-1/(2*n^2)+...) or 1-1/(2*n)+error where error<<1/n. Exponentiating shows our result is E*Exp[-1/(2*n)+error], now use what we know about n to show how close this comes to E.