You can describe the act of passing through a linear polarizer as a projection of the polarization degree of freedom of the first photon along the associated angle.
If for example the polarizer only lets through photons with the $\newcommand{\ket}[1]{\lvert #1\rangle}\ket +\equiv(\ket H+\ket V)/\sqrt2$ polarization, then the associated projector operator is $$\Pi_+\otimes I\equiv\ket+\!\langle+\rvert\otimes I,$$
where you have a projection of the polarization of the first photon, describing passing through the polarizer, and an identity on the state of the second photon, as this photon is unaffected by the polarizer.
To see the effect of this operation on your state, you just need to rewrite it in the appropriate basis. Straightforward calculations will lead you to write
$$\ket\Phi=\frac{1}{\sqrt2}(\ket{+_A}\otimes\ket{+_B} + \ket{-_A}\otimes\ket{-_B}).$$
In other words, the state is also maximally entangled when measured in the $\ket{\pm}\equiv(\ket{H}\pm\ket V)/\sqrt2$ basis (this is no coincidence: a maximally entangled state can be showed to be maximally entangled in any basis).
Passing through the polarizer you get (with some probability) the final state $\ket{+_A}\otimes\ket{+_B}$ (if you want to use as reference the polarization state at a $-45^\circ$ angle, just swap all $\ket{+}$s with $\ket{-}$s in the answer).
In conclusion, yes, the polarizer does "affect" the polarization of both photons because of their shared entanglement, and no the photons are not entangled anymore after the polarizer.
To be clear, this does not mean that actions performed on one of the photons somehow instantaneously affect the other photon.
Rather, it means that the outcome of the measurements on the two photons are correlated, so that the outcome of measuring the first photon changes our description of the second one, to reflect the newfound knowledge of the outcome of future measurements on it. So in this sense, the polarizer only affects the photon that passes through it, and the other photon is not affected by it.
