Is it possible to get Nth Fibonacci number in sublinear time?
Let's assume that you could store such large Fibonacci numbers. In that case, the  length of $n^{th}$ Fibonacci number is $\lfloor n \log_{10}\phi+1\rfloor$. That is the length of the $n^{th}$ Fibonacci number is $\Theta(n)$.

If you go by the naive method, then calculating the $n^{th}$ Fibonacci number would cost you $\Theta(n^2)$ time.

In comparison, matrix exponentiation will take $O(n^2\log n)$ time, assuming you use schoolbook long multiplication to multiply numbers.  (A slightly more refined analysis shows that it is in fact $\Theta(n^2)$, by summing a geometric series.) But if instead we use Karatsuba multiplication , the running time to compute the $n$th Fibonacci number using matrix exponentiation becomes $O(n^{1.585}\log n)$.  (Analogous to before, a slightly more refined analysis shows that the running time is in fact $\Theta(n^{1.585})$.) Thus, if you use sophisticated multiplication algorithms, matrix exponentiation would be better for large $n$ compared to the naive algorithm.

In short: which method is better depends on which multiplication algorithm used.