Here is a proof of the conjecture, a proof that also shows how to compute the integrals explicitly.
The proof is somewhat similar to David Speyer's approach, but instead of using multivariable residues, I will just shift a one-variable contour. Without loss of generality, $m>0$. Eliminating the trigonometric functions yields $I(m,n)=(-1/4\pi^2) J(m,n)$, where
$$ J(m,n) := \int_{|w|=1} \int_{|z|=1} \log(4+z+z^{-1}+w+w^{-1}) \, z^{m-1} w^{n-1} \, dz \, dw.$$
For fixed $w \ne -1$ on the unit circle, the values of $z$ such that $4+z+z^{-1}+w+w^{-1}$ are a negative real number and its inverse; let $g(w)$ be the value in $(-1,0)$. The function $\log(4+z+z^{-1}+w+w^{-1})$ of $z$ extends to the complement of the closed interval $[g(w),0]$ in a disk $|z|<1+\epsilon$. Shrink the contour $|z|=1$ like a rubber band so that it hugs the closed interval. The upper and lower parts nearly cancel, leaving
$$ J(m,n) = \int_{|w|=1} \int_{z=0}^{g(w)} -2\pi i \, z^{m-1} w^{n-1} \,dz \,dw,$$
with the $-2\pi i$ coming from the discrepancy in the values of $\log$ on either side of the closed interval. Thus
$$ J(m,n) = -\frac{2 \pi i}{m} \int_{|w|=1} g(w)^m w^{n-1} \, dw.$$
Next use the rational parametrization $(g(w),w) = \left( (1-u)/(u+u^2), (u^2-u)/(1+u) \right)$ that David Speyer wrote out. There is a path $\gamma$ from $-i$ to $i$ in the right half of the $u$-plane that maps to $|w|=1$ (counterclockwise from $-1$ to itself) and gives the correct $g(w)$. Integrating the resulting rational function of $u$ gives
$$ J(m,n) = -\frac{2 \pi i}{m} \left.(f(u) + r \log u + s \log(1+u))\right|_\gamma = -\frac{2 \pi i}{m} (f(i)-f(-i) + r \pi i + s \pi i/2),$$
for some $f \in \mathbf{Q}(u)$ and $r,s \in \mathbf{Q}$. This shows that $I(m,n)=a + b/\pi$ for some $a,b \in \mathbf{Q}$.
Examples: Mathematica got the answers wrong, presumably because it doesn't understand homotopy very well! Here are some actual values, computed using the approach above:
$$I(1,1) = \frac{1}{2} - \frac{2}{\pi}$$
$$I(2,1) = -1 + \frac{10}{3\pi}$$
$$I(20,10) = -\frac{14826977006}{5} + \frac{56979906453888224582}{6116306625 \pi}.$$
Let us give sufficient conditions on the growth rate of $N_l$ for the Fourier series to diverge.
We consider $p = N_l + 1$ and $q = \frac1M N_{l+1}$. For $p < |n| \leq q$ we have that $$ \hat{f}(n) = \sum_{j > l} 2^{-j}(1 - \frac{|n|}{N_j + 1}) \geq (1 - \frac{1}{M}) \sum_{j > l} 2^{-j} = (1 - \frac{1}M) \cdot 2^{-l}$$
we also have
$$ \hat{f}(n) - (1 - \frac1M) \cdot 2^{-l} \leq \frac1M \cdot 2^{-l}.$$
So
$$ \left| \sum_{p}^q \hat{f}(n) \cos(nt)\right| \geq (1 - \frac1M)2^{-l} \left|\sum_{p}^{q} \cos(nt)\right| - \frac{q-p}{M} 2^{-l} = (1 - \frac1M)2^{-l} |D_q(t) - D_p(t)| - \frac{q-p}{M} 2^{-l}.$$
Now you can use the divergence of the Dirichlet kernel (this follows from the fact that the $L^1$ norm of the Dirichlet kernel grows like $\log n$).
- Given $N_l$, we choose $q$ such that $\| D_{q} - D_{N_{l}+1}\|_{L^1} \geq 2^{l}$.
- Next we choose $M$ such that $M > 2 (q - N_{l} - 1)$.
- We choose $N_{l+1} > M q$.
- repeat with $l \mapsto l+1$.
Then we have
$$ \|S_q(f) - S_p(f)\|_{L^1} \geq \frac12 ( 1 - 2^{-l}) $$
for the $p,q$ associated to each $l$, showing that the Fourier series cannot converge in $L^1$.
Best Answer
$$\int_0^{2\pi} \exp(int) \exp(\cos(t))\; dt = \int_{-\pi}^{\pi} \cos(n t) \exp(\cos(t))\; dt = 2 \pi I_n(1)$$ where $I_n$ is a modified Bessel function of the first kind and thus $$I_n(1)=\frac12\sum_{k\geq0}\frac1{4^kk!(n+k)!}.$$