The answer is 6.
Let $a_n = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n}$. Firstly, we find that
$$a_{n+1} - a_n = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)} > 0.$$
This shows the sequence is strictly increasing. Thus, we easily find that
$$a_{1008} > a_3 = \frac{37}{60} > 0.6.$$
Secondly, we can prove by induction that $a_n \le 0.7 - \frac{1}{4n}$ for any $n \ge 3$. The base case $a_3 = \frac{37}{60} \le 0.7 - \frac{1}{12}$ holds. Assuming this holds for some $n = k$, we find that
$$a_{k+1} = a_k + \frac{1}{(2k+1)(2k+2)} \le 0.7 - \frac{1}{4k} + \frac{1}{(2k+1)(2k+2)}.$$
Now,
$$\begin{align*} & 0.7 - \frac{1}{4k} + \frac{1}{(2k+1)(2k+2)} \le 0.7 - \frac{1}{4(k+1)} \\ \Leftrightarrow \quad & \frac{1}{(2k+1)(2k+2)} \le \frac{1}{4k(k+1)} \\ \Leftrightarrow \quad & 4k(k+1) \le (2k+1)(2k+2) \\ \Leftrightarrow \quad & 0 \le 2k+2. \end{align*}$$
This clearly holds. By induction, we have $a_n \le 0.7 - \frac{1}{4n} < 0.7$ for any $n \ge 3$. In particular, $a_{1008} < 0.7$. Therefore, the first digit of $a_{1008}$ after the decimal point is 6.