After noting that $x \to e^{ixp}$ is an eigenvector (with eigenvalue $p$) of the endomorphism $\hat p:u \to \frac{1}{i} \frac{\partial u}{\partial x}$, my teacher goes one step beyond and states that $x \to \int dp\Phi(p)e^{ixp}$ is also an eigenvector (with eigenvalue $p$) of the same endomorphism $\hat p$ using the linearity argument.
I've understood the linearity argument. However, I don't understand how, "formally", we can conclude this result since $$\hat p (x \to \int dp\Phi(p)e^{ixp}) = x \to\frac{1}{i} \frac{\partial }{\partial x}\int dp\Phi(p)e^{ixp}\\ =x\to\frac{1}{i}\int dp\Phi(p)\frac{\partial}{\partial x}e^{ixp}=x\to\int dp\Phi(p)pe^{ixp}$$ and this last term is from my point of view (I may be wrong...!) different from $p(x\to\int dp\Phi(p)e^{ixp})$
I already asked in physics.SE (since it departs from a Quantum Mechanics problem), I am confused about it.
You have badly misunderstood your teacher, or he has misled you, badly.
Indeed, the momentum operator is linear, $$ \hat p \equiv \int\!\! dx ~|x\rangle \frac{\partial_x}{i} \langle x| , \implies \\ \langle x|\hat p|u\rangle \equiv \hat p _x ~u(x)= \frac{\partial_x u}{i}, $$ so that $u(x)= e^{ixp}$ is an eigenvector of it, with eigenvalue $p$.
The statement that $ u(x)= \int dp\Phi(p)e^{ixp}$ is also an eigenvector of it is flat wrong. Indeed, by linearity, $$\hat p_x ~ \int\!\! dp~\Phi(p)e^{ixp} = \frac{\partial_x}{i} \int\!\! dp~\Phi(p)e^{ixp} \\ = \int dp~\Phi(p)~p~e^{ixp},$$ and and you are right you may not take the dummy variable out of the integral, so that integral is not an eigenvector of the momentum operator in the x-representation.
You might discuss with your instructor what he had in mind, but, almost certainly, not the false statement you are positing. The statement is only true for $\Phi(p)=\delta (p-p_0)$, in which case the eigenvalue would be $p_0$, and "almost" correct with the δ-function supplanted by a very sharply peaked function.
Addendum post OP's comment
Indeed, if u(x) corresponds to a sharply peaked $$\Phi (p-p_0)=\frac{e^{-(p-p_0)^2/\epsilon }}{\sqrt{\pi \epsilon}} \\ \lim_{\epsilon\to 0} \Phi(p)=\delta (p-p_0), $$ you can see from your integral that $$ \hat p_x u(x)= p_0 u(x) + \sqrt{\frac{\epsilon}{\pi}}e^{ixp_0}\int \!\!dy ~y e^{-y^2} e^{ixy\sqrt{\epsilon}} $$ so the eigenvalue equation is corrected by an insignificant term of $O(\sqrt \epsilon)$.