The intersection numbers $H^4 =1,\; H^3\cdot E = 0,\; H^{2}\cdot E^2 = -2,\; H\cdot E^3 = -6$ are correct. The last one $E^4 = H^0\cdot E^4 = -18$ is wrong.

A way to compute $E^4$ is the following. The blow-up $X$ of a smooth quadric surface $Q\subset\mathbb{P}^4$ is isomorphic to the blow-up of a smooth $4$-dimensional quadric in a point. Let us call $Y$ this blow-up, and let $E_p\subset Y$ be the exceptional divisor. Then $E_p^4 = -1$. On the other hand $E_p$ is the strict transform through the blow-up map $\epsilon:X\rightarrow \mathbb{P}^4$ of the $3$-plane spanned by $Q$. Then $E_p = H-E$, and
$$E_p^4 = H^4-4H^3\cdot E+6H^2\cdot E^2-4H\cdot E^3+E^4.$$
Using the first four intersection numbers we get
$$-1 = E_p^4 = 1-12+24+E^4,$$
and finally $E^4 = -14$. Summing up the intersection numbers you are looking for are:
$$H^4 =1,\; H^3\cdot E = 0,\; H^{2}\cdot E^2 = -2,\; H\cdot E^3 = -6,\; E^4 = -14.$$
Another way to see this is the following:
$$E^4 = -deg(Q)s_2 = -2s_2,$$
where $s_2$ is the second Segre class of the conormal bundle $N_{Q/\mathbb{P}^4}^{\vee} = \mathcal{O}_{Q}(-1)\oplus\mathcal{O}_{Q}(-2)$. The chern classes of this bundle are $c_1 = -3$ and $c_2 = 2$. Therefore $s_1 = 3$ and $s_2 = s_1^2-c_2 = 7$. Finally you get again $E^4 = -2s_2 = -14$.