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.
Instead, we need that there is a $c_i$ such that $c_{i-1}$ is before it and $c_{i+1}$ is after it, or the other way around.**
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, which thus practically act on $m=\frac{3n}k$ elements.
Every cycle $C$ of length $k$ either has three groups that contain three consecutive vertices from $C$, 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).
For example, $c_1$ and $c_2$ are in the first group, with $c_1<c_2$, while $c_3$ is in the second group.
In this case, all we need is a permutation where the second group is after the first group, then $c_2$ will be between its neighbors.
Similarly, if $c_1, c_2$ and $c_3$ are all in different groups, then we need that the group of $c_2$ is between the other two groups.
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.
In more detail, there are $m^3$ ordered triples, each permutation has probability $\frac 16$ to work for each triple, thus if we take $t$ independent permutations, the chance for any ordered tripe that none of the permutations work for it is $(1-\frac 16)^t$ and taking the union bound we need $m^3(1-\frac 16)^t<1$.