Short answer: $H^i(\mathbb{Q}/\mathbb{Z}) = 
\begin{cases}
\mathbb{Z}, i = 0,\\
0, i \equiv 1 \mod 2,\\
A, i\equiv 0 \mod 2, i>1. 
\end{cases}$

Where $A$ fits into a short exact sequence $\mathbb{Z}\to A\to H^2(\mathbb{Q})$. 

Explanation: 

We can use the short exact sequence to compute the cohomology of $\mathbb{Q}$ from the cohomology of $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Z}$. 

If we look at the spectral sequence we see that only the first two rows are non-zero since the $(p, q)$ entry is $H^p(\mathbb{Q}/\mathbb{Z}; H^q(\mathbb{Z}; \mathbb{Z}))$ which is non-zero only if $H^q(\mathbb{Z})\neq 0$ only if $q = 0, 1$. 

Thus we end up with a long exact sequence of cohomology groups
$$0 \to H^1(\mathbb{Q}/\mathbb{Z})\to H^1(\mathbb{Q})\to H^0(\mathbb{Q}/\mathbb{Z})\to \dots \to H^i(\mathbb{Q})\to H^{i-1}(\mathbb{Q}/\mathbb{Z}) \to H^{i+1}(\mathbb{Q}/\mathbb{Z})\to H^{i+1}(\mathbb{Q})\to\dots.$$ 

Now $\mathbb{Q}$ has cohomological dimension 2. So we obtain $H^i(\mathbb{Q}/\mathbb{Z}) = H^{i+2}(\mathbb{Q}/\mathbb{Z})$ for $i>2$. 

For the low dimensional cases since $\mathbb{Q}/\mathbb{Z}$ is torsion we see $H^1(\mathbb{Q}/\mathbb{Z})$ vanishes. 

Also $H^1(\mathbb{Q})$ vanishes so we get a short exact sequence $0\to H^0(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q})\to 0$. Hence the even terms. 

We also see that a portion of the long exact sequence is: 
$$\dots\to H^{1}(\mathbb{Q}/\mathbb{Z}) \to H^{3}(\mathbb{Q}/\mathbb{Z})\to H^{3}(\mathbb{Q})\to \dots.$$
So $H^3(\mathbb{Q}/\mathbb{Z})$ vanishes. 

Unfortunately, I do not understand $H^2(\mathbb{Q})$, let alone an extension of it by $\mathbb{Z}$. I believe the universal coefficient theorem shows that $H^2(\mathbb{Q}\cong $ Ext$(\mathbb{Q}, \mathbb{Z})$ which is described in some detail on page 5 of these notes: http://math.jhu.edu/~jmb/note/torext.pdf