From the problem statement, we can have an initial $P_1, V_1, \text{ and } T_1$ with no gas or heat heat flowing from it, and the model of the Ideal gas Law and the Universal Gas Constant so that: $$P_1 \times V_1 = n_1 \times R \times T_1$$ And quoting from the Reference 2 Universal Gas Constant, then
The universal gas constant, also known as the molar or ideal gas constant, is:
$$ R^* = 8.3144621\text{ }(75)\text{ } \frac{J}{\text{mol} \times K} $$
The gas constant for a particular gas is
$$ R = \frac{R^*}{m} $$
where m is the molecular weight of the gas. For a mixture, the "molecular weight" is a weighted mean of the molecular weights of the components:
$$m=\left( \frac{f_1}{m_1} + · · · + \frac{f_n}{m_n} \right)^{-1}$$
$$\text{where } m_1, · · ·, m_n$$
are the molecular weights of the n gases,
$$\text{and } f_1, · · ·, f_n$$
are their masses relative to the total mass of the mixture. The gas constant for dry air is
$$R_d = 287 \frac{J}{K \times kg}$$
The problem statement, "They don't create any chemical reaction (it would be O², N² and CO²)" is not precisely stated dry air. This is a difficulty with the problem statement which may affect the answer quality, because when $R_d$ is used in the ideal gas law it may different from $R_\text{1 for the question}$ in the context of the question presentation.
Technically we do not have everything required to calculate the R value for the gas mixture in the room. Hence, the answer needs to be conceptual about how to proceed (and I am amending the question a little with more detail to make it solvable), and then the actual Gas Constant for the room can perhaps be obtained later to follow the similar procedure to arrive at the desired answer, starting with the assumption that all the gasses are dry air and using the corresponding gas constant $R_d$ as:
$$R_d = 287 \frac{J}{K \times kg}$$ $$\text{and initially } P_1 \times V_1 = n_1 \times R_{1\text{ d}} \times T_1$$ $$\text{and finally } P_2 \times V_2 = n_2 \times R_{2\text{ d}} \times T_2$$
Now what happens with the next event in the question statement, "What happens when n suddenly changes?" The usual meaning of "suddenly changes" is a model of an Adiabatic Process, and from this reference:
If the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, its temperature does not change; this is also called a Joule expansion.
The question here states: "You can picture a compressed gas container leaking n moles of gas". When I imagine this, I see a picture of a gas container at very high pressure compared to its environment, and just a little bit of gas leaking. This is not the scenario where the gas container is almost empty and at equilibrium with the atmosphere that it leaks to since that entity is completely outside of the problem statement.
Ideal Gas Thermodynamics: Specific Heats, Isotherms, Adiabats has the following illustration which shows what I envision from the problem statement. Some gas is let out of the gas container quickly resulting is a situation similar to that pictured here. When the high pressure gas from the container is let out, the volume of the gas inside the container and that outside of the container suddenly increases without additional significant release of heat (beyond that already in the gas that is leaked) to the environment.
Also, since this is a small and sudden leak, the work can be estimated as a relatively constant pressure times a small change in volume as:
$$\delta W_{\text{work from the container}}= P \times \delta V_{\text{volume of container and environment}}$$ $$=\delta n \times R_d \times T_{\text{Adiabatic constant in the gas container}}$$
From this equation one can see that the average internal energy of each molecule does not change, and rather it is the loss of molecules $\delta n$ during the gas release from the container that changes the complete energy stored in the container. The temperature $T$ is related to the average energy of each ideal gas molecule in the gas container, not to how many gas molecules there are. Thus, $T$ is constant, and total energy loss in the container is just the average energy per molecule times the number of molecules.
The initial temperature $T_1$ of course needs to be independent of the environment temperature that the gas is leaking to $T_\text{environment}$ because this is envisioned as a high-pressure gas container with nominal influence of the environment that it is leaking to (where the details of its temperature and so forth are not given in the question statement).
So finally,
$$\delta W_{\text{work from the container to the environment}}= $$ $$= P \times \delta V_{\text{volume of container and environment}}$$ $$=\delta n \times R_d \times T_{\text{Adiabatic constant in the gas container}}$$
implies:
$$\delta \left(P \times V_{\text{container}}\right) = \delta P \times V_{\text{fixed size gas container}}$$ $$=\delta \left( n \times R_d \times T_{\text{Adiabatic constant for Joule heating in the gas container}}\right)$$
$$=\delta \left( n_2-n \right) \times R_d \times T_{\text{Adiabatic constant in the gas container}}$$
Just like a natural gas container of high pressure gas, the pressure starts out high and slowly decreases according to how much gas
$$\left( n_2- n \right)$$
is suddenly used, while the natural gas container volume remains constant.
Thus, contrary to the initial perception that: "According to my intuition both Pressure P and Temperature T would increase." **It is quite different. Think like it is a metal high pressure gas canister for a fixed volume of gas, like for soda production. A release of gas decreases pressure and the volume remains fixed. Also, Temperature does not increase. Rather, because this is Joule work that the gas performs, and each molecule inside the container retains its average energy, the Temperature remains the same.