Let $(E,\mathcal E,\mu)$ be a probability space, $(X_n)_{n\in\mathbb N_0}$ be an $(E,\mathcal E)$-valued stationary time-homogeneous Markov chain with initial distribution $\mu$ and $$A_nf:=\frac1n\sum_{i=0}^{n-1}f(X_i)\;\;\;\text{for }n\in\mathbb N\text{ and }f\in\mathcal L^1(\mu).$$
Assume $f:E\to[0,\infty)^3$ is $\mathcal E$-measurable and $$\sqrt n\left(A_nf_i-\int f_i\:{\rm d}\mu\right)\xrightarrow{n\to\infty}\mathcal N_{0,\:\sigma_i^2}\;\;\;\text{in distribution for all }i=1,2,3.\tag1$$
Can we conclude that $\sqrt n\left(A_nf-\int f\:{\rm d}\mu\right)$ converges to a multidimensional normal distribution?
In general this implication is wrong, while the converse is true by the continuous mapping theorem.
The claim is true, even in a way more general setting.
Assume $H$ is a $\mathbb R$-Banach space and $A_nf$ is any estimator of $\mu f$ for $f\in\mathcal L^1(\mu)$ which is linear in $f$ and satisfies $$A_nf\xrightarrow{n\to\infty}\mathcal N_{0,\:\sigma^2(f)}\tag2$$ for all $f$ belonging to a subclass $\mathcal C$ of $\mathcal L^1(\mu)$. If $f:E\to H$ is $\mathcal E$-measurable and $$\varphi\circ f\subseteq\mathcal C\;\;\;\text{for all }\varphi\in H',$$ then $$\varphi\circ\sqrt n(A_nf-\mu f)=\sqrt n(A_n(\varphi\circ f)-\mu(\varphi\circ f))\xrightarrow{n\to\infty}\mathcal N_{0,\:\sigma^2(\varphi\:\circ\:f)}\;\;\;\text{for all }\varphi\in H'\tag3$$ and hence $$\sqrt n(A_nf-\mu f)\xrightarrow{n\to\infty}X$$ for some $H$-valued random variable $X$ with $$\mathcal L(\varphi\circ X)=\mathcal N_{0,\:\sigma^2(\varphi\circ f)}\;\;\;\text{for all }\varphi\in H'.\tag4$$ The claim in the question is immediate from that.