Clearly $I_i$ is an additive subgroup of ${\bf Z}$, hence every $I_i$ corresponds uniquely to a positive divisor $d_i$ of $168=2^3\cdot 3\cdot 7$. Also we need $d_1|d_2|\cdots |d_k$. For $k=1$ we have $\tau(2^3\cdot 3\cdot 7)=4\cdot 2\cdot 2=16$. For $k\ge 2$ we have $I_1=(2^{a_1}3^{b_1}7^{c_1}),\cdots,I_k=(2^{a_k}3^{b_k}7^{c_k})$ where $a_1\le\cdots \le a_k\le 3,b_1\le \cdots \le b_k\le 1,c_1\le \cdots \le c_k\le 1$. We can count the number of sequences $a,b,c$ separately and then we multiply. The sequences $a$ correspond bijectively to sequences $\alpha_1,\cdots,\alpha_k$ with $\alpha_i\ge 0$ and $\sum \alpha_i \le 3$. With stars-and-bars this is the number of words with 3 zeroes and $k$ ones, where for example the word $0110110$ corresponds with $(\alpha_1,\cdots,\alpha_4)=(1,0,1,0)$ [count the number of zeroes separated by the ones and omit any zeroes to the far right] and $(a_1,\cdots,a_4)=(1,1,2,2)$. Hence we have ${{k+3}\choose 3}$ sequences $a$, we have $k+1$ sequences $b$, we have $k+1$ sequences $c$ and the total number of $k$-chains becomes $(k+1)^2 {{k+3}\choose 3}$.