Complexity of k-way merge with log n arrays of size n / log n
First of all, notice that whatever you were doing, the complexity up to this point can not possibly be $o(n \log n)$. If that were true, we could merge each sorted subsequence in time $O(n \log \log n)$, violating the well known bound of $\Omega(n \log n)$ for the worst case of a general sorting algorithm. In fact, $$\Theta\left(n \log \frac{n}{\log n} \right) = \Theta(n \log n - n \log \log n) = \Theta (n \log n)$$

Now, let's analyze the complexity of what you are proposing. In order to visualize all the merging that's going on, imagine inserting each sorted subsequence into a FIFO queue. At each step, we take the first two sequences of the queue, merge them, and put the result at the back of the queue. We take note of the sequence that is initially in front of the queue, and define a round as the set of all merging operations that happen between two subsequent appearances of the initial sequence at the front of the queue.

Observe that each round splits in half the queue size: let's say there are $r$ rounds, with $r \in \Theta(\log \log n)$. Let $\alpha = \frac{n}{\log n}$. Before the $h-$th round begins, the queue contains $\frac{n}{2^h \alpha}$ subsequences, each of which has length $2^h \alpha$. Therefore, the complexity of round $h$ is:

$$
\Theta\left( \frac{n}{2^{h+1} \alpha} \left( 2^{h+1} \alpha \log (2^{h+1} \alpha) \right) \right) = \Theta \left( n \log (2^{h+1} \alpha) \right)
$$

This yields an overall complexity of:

$$
\Theta\left( \sum_{h=0}^{r-1} n \log (2^{h+1} \alpha) \right) =
\Theta\left( nr \log \alpha + n\sum_{h=0}^{r-1} \log 2^{h+1} \right ) =
\Theta\left( nr \log \alpha + nr^2 \right )
$$

Now, remembering that:

$$
\Theta \left( \log \alpha \right) = \Theta \left( \log n - \log \log n \right) = \Theta(\log n)
$$

We conclude:

$$
\Theta\left( nr \log \alpha + nr^2 \right ) = \Theta(n \log n \log \log n)
$$