Thermodynamics of chemical reactions "at constant pressure"
You have correctly stated the pressure of the system is $$p = \frac {(n_1 + n_2)RT}{V}$$
But at the end of your question you incorrectly imply the volume is constant and the pressure changes. That is not true. Constant pressure means the external constant pressure ( like the atmospheric one ) keeps the system pressure constant. Imagine there is a massless, frictionless piston, ensuring $p_\mathrm{ext} = p_\mathrm{sys}$. It is similar as a thermostatic water bath keeps the system temperature constant.
So less confusing is to rewrite the ideal gas state equation :
$$V = \frac {(n_1 + n_2)RT}{p}$$
There are 4 variables: $p, V, T, n_\mathrm{tot}=n_1 + n_2$.
But there are just 3 degress of freedom = 3 independent variables.
Either $V, n_\mathrm{tot}, T$ are given, and $p=f(V, n_\mathrm{tot},T)$,
either $p, V, T$  are given, and $n_\mathrm{tot}=f(p,V,T)$,
either $p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)$,
either $p, n_\mathrm{tot}, V$  are given, and $T=f(p,n_\mathrm{tot},V)$.
We have already given 2 variables: $p, T$, with $n_\mathrm{tot},V$ remaining free.
But there is the equilibrium relation
$$K_p = \frac {p_{\ce{N2O4}}} {({p_{\ce{NO2}} )}^2}=\frac {p \cdot ( 1 - x_{\ce{NO2}}) } {({p \cdot x_{\ce{NO2}} )}^2}$$
$$K_p \cdot p \cdot {({x_{\ce{NO2}} )}^2} + x_{\ce{NO2}} - 1 = 0$$
Solving the quadratic equation we would finally get $n_\mathrm{tot}$, so all 3 degrees of freedom are saturated with given values for $p, T, n_\mathrm{tot}$.
So in our case, the following applies:
$p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)=\dfrac {n_\mathrm{tot}RT}{p}$