yes, your method is absolutely correct until this point. Just progress a bit more. See :
2^k T(1) = n T(1) = n
Now how many n's are added in the expression. If you count it, you got 3 n's for T(n/8). You could also check you got 2 n's for T(n/4). i.e you get k n's to add up for T(n/ 2^k)
Thus,
2 ^ k T(1) + n + n + n + ... = 2^k T(1) + k n = n T(1) + log n . n [ since, 2^k=n => k=log n]
= n + n log n = n log n