$T(n) = \sum_{i=1}^{n} \sum_{j=1}^{i} \sum_{k=1}^{j} \sum_{l=1}^{k} c$
( Taking cost of printf as $c$)
$= c\sum_{i=1}^{n} \sum_{j=1}^{i} \sum_{k=1}^{j} k $
$= c\sum_{i=1}^{n} \sum_{j=1}^{i} \frac{j.(j+1)}{2} $
$= d \sum_{i=1}^{n} \sum_{j=1}^{i} (j^2 + j)$
$(d = c/2)$
$= d \sum_{i=1}^{n} \frac{i.(i+1)(2i+1)}{6} + \frac{i.(i+1)}{2})$
$= e \sum_{i=1}^{n} 2i^3 + 3i^2 + i + 3i^2 + 3i $
$(e = d/6)$
$= e \sum_{i=1}^{n} 2i^3 + 6i^2 + 4i $
Well, sum of the cubes of first $n$ natural numbers is $\left({\frac{n.(n+1)}{2}}\right)^2 = \Theta\left (n^4\right)$
So, our answer will also be $\Theta\left( n^4\right)$ due to the $i^3$ term going from $i=1..n$