Counting might help

First notice that $210=2\cdot 3\cdot 5\cdot 7$

Now let $a=2^{x_1}3^{y_1}5^{z_1}7^{\omega_1} \, , b=2^{x_2}3^{y_2}5^{z_2}7^{\omega_2} \, , c=2^{x_3}3^{y_3}5^{z_3}7^{\omega_3} \, , d=2^{x_4}3^{y_4}5^{z_4}7^{\omega_4}$

So $\quad a\cdot b\cdot c\cdot d =2^{x_1}3^{y_1}5^{z_1}7^{\omega_1}\cdot 2^{x_2}3^{y_2}5^{z_2}7^{\omega_2} \cdot 2^{x_3}3^{y_3}5^{z_3}7^{\omega_3} \cdot 2^{x_4}3^{y_4}5^{z_4}7^{\omega_4}$

$\begin{aligned} \\ 0&\leq x_1+x_2+x_3+x_4&\leq 1 \\ 0&\leq y_1+y_2+y_3+y_4&\leq 1 \\ 0&\leq z_1+z_2+z_3+z_4&\leq 1 \\ 0&\leq \omega_1+\omega_2+\omega_3+\omega_4&\leq 1 \end{aligned}$

Each of these equations have $4$ positive integer solutions meaning $abcd=210$ has $4\times 4\times 4\times 4=4^4$ positive integer solutions.

Now for all integer solutions we must consider negatives. We can arbitrarily assign the signs to any three positive integers and only the last one will determine if the result is positive or negative. We can assign the signs in $2^3=8$ ways.

So the number of integer solutions is $2^3\cdot 4^4=2^3\cdot 2^8=2^{11} \implies 2n=2\cdot 11=\boxed{22}$