For $n>1$, the natural guess would be that $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=I_{Y/X}^n$, and it's true in this case. For $n<0$, we have $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=\mathcal{O}_X$.
To show these claims, we'll work locally on $X$. Assume $X=\operatorname{Spec} A$ is affine and $Y$ is cut out by the ideal $I_Y=(f_1,\cdots,f_r)$. Then we get a surjection $A^r\to (f_1,\cdots,f_r)$ which turns in to a surjection of the graded algebras $\operatorname{Sym}(A^r)\to \bigoplus_{m\geq 0} I_Y^m$ corresponding to the closed immersion $\widetilde{X}\hookrightarrow \Bbb P(A^r)$. Here, the exceptional divisor $E$ corresponds to the line bundle $\mathcal{O}_{\Bbb P(A^r)}(-1)|_{\widetilde{X}}$.
Next, via the restriction map $\mathcal{O}_{\Bbb P(A^r)}(m) \to \mathcal{O}_{\Bbb P(A^r)}(m)|_{\widetilde{X}}$ and the canonical isomorphism $\mathcal{O}_{\widetilde{X}}(1)\cong \mathcal{O}_{\widetilde{X}}(-E)$, we see that if $\mathcal{O}_{\Bbb P(A^r)}(n)$ is globally generated, then $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n)\to \pi_*\mathcal{O}_{\widetilde{X}}(-nE)$ will be surjective and thus an isomorphism of line bundles. By identifying $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n) = \operatorname{Sym}^n(A^r)$ and $\pi_*\mathcal{O}_{\widetilde{X}}(-nE)=I^n$, we see that when the global generation condition is satisfied, then we have an isomorphism $\mathcal{O}_{\widetilde{X}}(-nE)\cong I_Y^n$.
By Serre vanishing, this is always the case for any $X,Y$ assuming that $n>>0$. In our case, the argument from the linked post shows that in fact $\mathcal{O}_{\Bbb P(A^r)}(1)$ is globally generated, so $\mathcal{O}_{\Bbb P(A^r)}(n)$ is globally generated for all $n>0$. So we get our claimed isomorphism $\pi_*\mathcal{O}_\widetilde{X}(-nE)\cong I^n_Y$.
For $n<0$, after tensoring the natural exact sequence $0\to \mathcal{O}_\widetilde{X}(nE) \to \mathcal{O}_X\to\mathcal{K} \to 0$ by $\mathcal{O}_\widetilde{X}(-nE)$ we get the sequence $0\to \mathcal{O}_\widetilde{X}\to \mathcal{O}_\widetilde{X}(-nE)\to \mathcal{K}(-nE)\to 0$ (here $\mathcal{K}$ is the structure sheaf of a thickening of $E$). If we prove that $\pi_*\mathcal{K}(-nE)$ has no global sections, then it's the zero sheaf as $X$ is affine, and this would imply $\mathcal{O}_X=\pi_*\mathcal{O}_\widetilde{X}\to \pi_*\mathcal{O}_\widetilde{X}(-nE)$ is an isomorphism.
As $\mathcal{O}_\widetilde{X}(-nE)\cong \mathcal{O}_\widetilde{X}(n)$, we see that it's a negative line bundle, and after restriction to any projective subvariety of $\Bbb P(A^r)$ it will still be a negative line bundle and therefore have no sections. Taking the fiber $\Bbb P(A^r)_y$ for $y\in Y$, we see that this is a projective variety, so there are no sections of $\mathcal{O}_\widetilde{X}(n)$ in the fiber direction along $E\to Y$. But any global section of $\pi_*\mathcal{K}(-nE)$ would come from such a global section of $\mathcal{O}_\widetilde{X}(n)$ because $\mathcal{O}_\widetilde{X}(n)\to \mathcal{K}(-nE)$ is surjective and would remain this way after restricting to $E$. So there cannot be any global sections and we've shown that $\pi_*\mathcal{O}_\widetilde{X}(-nE)=\mathcal{O}_X$ for $n<0$.
Best Answer
To compute the image you can replace $I_p$ by $\mathcal{O}_X$ and the copies of $E$ by $I_{D_i}$. Then your question becomes: what is the image of $$ I_{D_1} \oplus I_{D_2} \oplus I_{D_3} \oplus I_{D_4} \to \mathcal{O}_X. $$ The image of the direct sum of ideal sheaves is, of course, the sum of the corresponding ideals, i.e., the ideal of the corresponding intersection of schemes. So, it proves that the image you are interested in is $I_Z$, where $$ Z = D_1 \cap D_2 \cap D_3 \cap D_4. $$ In particular, if you want to prove it is $I_p$, you need to check that the scheme-theoretic intersection of your quintic curves is the point $p$.