The answer is $13725$.
We first claim that if $n$ is odd, then $\prod_{b=1}^{n} (1+e^{2\pi i ab/n}) = 2^{\gcd(a,n)}$. To see this, write $d = \gcd(a,n)$ and $a = da_1$, $n=dn_1$ with $\gcd(a_1,n_1) = 1$. Then
$a_1, 2a_1,\dots,n_1 a_1$ modulo $n_1$ is a permutation of $1,2,\dots,n_1$ modulo $n_1$, and so $\omega^{a_1},\omega^{2a_1},\dots,\omega^{n_1 a_1}$ is a permutation of $\omega,\omega^2,\ldots,\omega^{n_1}$; it follows that for $\omega = e^{2\pi i/n_1}$,
$$\prod_{b=1}^{n_1} (1+e^{2\pi i a b/n}) = \prod_{b=1}^{n_1} (1+e^{2\pi i a_1 b/n_1}) = \prod_{b=1}^{n_1} (1+\omega^b).$$
Now since the roots of $z^{n_1}-1$ are $\omega,\omega^2,\ldots,\omega^{n_1}$, it follows that
$z^{n_1}-1 = \prod_{b=1}^{n_1} (z-\omega^b)$. Setting $z=-1$ and using the fact that $n_1$ is odd gives $\prod_{b=1}^{n_1} (1+\omega^b) = 2$.

Finally,
$\prod_{b=1}^{n} (1+e^{2\pi i ab/n}) = (\prod_{b=1}^{n_1} (1+e^{2\pi i ab/n}))^d = 2^d$, and we have proven the claim.

From the claim, we find that
$$\begin{align*} &\log_2 \left( \prod_{a=1}^{2015} \prod_{b=1}^{2015} (1+e^{2\pi i a b/2015}) \right) \\ &= \sum_{a=1}^{2015} \log_2 \left(\prod_{b=1}^{2015} (1+e^{2\pi i a b/2015}) \right) \\ &= \sum_{a=1}^{2015} \gcd(a,2015). \end{align*}$$
Now for each divisor $d$ of $2015$, there are $\phi(2015/d)$ integers between $1$ and $2015$ inclusive whose $\gcd$ with $2015$ is $d$. Thus
$$\sum_{a=1}^{2015} \gcd(a,2015) = \sum_{d|2015} d\cdot \phi(2015/d).$$
We factor $2015 = pqr$ with $p=5$, $q=13$, and $r=31$, and calculate
$$\begin{align*} &\sum_{d|pqr} d\cdot \phi(pqr/d) \\ &= 1 \cdot (p-1)(q-1)(r-1) + p \cdot (q-1)(r-1) \\ &\quad + q\cdot (p-1)(r-1) + r\cdot (p-1)(q-1) + pq \cdot (r-1) \\ & \quad + pr\cdot (q-1) + qr\cdot (p-1) + pqr \cdot 1 \\ &\quad = (2p-1)(2q-1)(2r-1). \end{align*}$$
When $(p,q,r) = (5,13,31)$, this is equal to $13725$.