Update 2015.08.07: This proof had an error, as pointed out by Turbo, but hopefully not fatal, below I tried to fix it.
Here is a modest improvement on the upper bound that shows $d=O(n\log\log n)$.
The base of the construction is the following. Order the vertices as $v_1,\ldots,v_n$. Fix $2n$ independent vectors from $\mathbb F_2^d$ and assign two vectors, $\ell_i$ and $r_i$, to each vertex $v_i$. Assign to the edge $v_iv_j$ the vector $r_i+\ell_j$ if $i<j$ and $\ell_i+r_j$ if $i>j$. Notice that this gives a nonzero sum for a cycle $c_1c_2\ldots c_k$ if and only if the length of the cycle is even and every eventh vertex comes before (or after) every oddth vertex. <- This is false, but luckily we won't need it. Therefore, it has a good chance to work for long cycles.
Our strategy will be to take $t$ random permutations (with $t\cdot 2n$ independent vectors) and take the sums on each edge to take care of all cycles longer than $k$. The standard probabilistic approach with $\approx k\log n$ further independent vectors takes care of all cycles whose length is less than $k$. This in total gives $d\approx t\cdot n + k\log n$.
Claim. $t\approx \log \frac nk$ random permutations are enough to take care of all cycles longer than $k$.
After this we can choose $k=\frac {n\log \log n}{\log n}$ to get the desired $d=O(n\log\log n)$ bound.
Proof of the claim (new). Divide $n$ into groups of size $\frac k3$. Fix an order inside each group, this will be the same in each permutation. Every cycle of length $k$ either has three groups that contain three consecutive vertices, or two groups, one of which contains two of three consecutive vertices and the other group the third one (this latter case is better for us). Therefore, it is enough to show that for $m=\frac{3n}k$ elements there are $O(\log m)$ permutations such that for any three elements $a,b,c$ there is a permutation in which $a<b<c$. A standard probabilistic argument shows that $O(\log m)$ random permutations work.