$\newcommand{\al}{\alpha}$This inequality is false, at least if $\al\ne1$. Indeed, suppose that $C_1(x)+C_2(x)=1$ for all $x>1$, \begin{equation} \ell(x)=e^{b\sqrt{\ln x}} \end{equation} for some real $b$ and all $x>1$, and $f(t)=\ln^2 t$ for all $t>2$.

Then, reasoning as in your post, for all large enough $n$ we have \begin{equation} P(|X|>a_n)\asymp\frac1{nf(n)}\frac{\ell(a_n)}{\ell(n)} =\frac1{n\ln^2 n}\frac{\ell(a_n)}{\ell(n)}, \end{equation} $a_n=n^{1/\al+o(1)}$, \begin{equation} \frac{\ell(a_n)}{\ell(n)}=\exp\{b(\sqrt{1/\al}-1+o(1))\sqrt{\ln n}\}. \end{equation} Letting now $b=1$ if $\al\in(0,1)$ and $b=-1$ if $\al\in(1,2)$, we see that for all large enough $n$ \begin{equation} \frac{\ell(a_n)}{\ell(n)}>\ln n \end{equation} and hence \begin{equation} \sum_n P(|X|>a_n)=\infty, \end{equation} whereas $\int_1^\infty \frac{dt}{t f(t)}<\infty$, so that the inequality \begin{equation} \sum_{n=1}^\infty P\big( |X| > a_n \big) \le \tilde C \int_1^\infty \frac{dt}{t f(t)} \end{equation} cannot hold for any real $\tilde C$.

$\newcommand{\al}{\alpha}$This inequality is false, for any $\al\in(0,2)$. 

Indeed, consider first the case $\al\ne1$. Suppose that $C_1(x)+C_2(x)=1$ for all $x>1$, \begin{equation} \ell(x)=e^{b\sqrt{\ln x}} \tag{1}\label{1} \end{equation} for some real $b$ and all $x>1$, and $f(t)=\ln^2 t$ for all $t>2$.

Then, reasoning as in your post, for all large enough $n$ we have \begin{equation} P(|X|>a_n)\asymp\frac1{nf(n)}\frac{\ell(a_n)}{\ell(n)} =\frac1{n\ln^2 n}\frac{\ell(a_n)}{\ell(n)}, \end{equation} $a_n=n^{1/\al+o(1)}$, \begin{equation} \frac{\ell(a_n)}{\ell(n)}=\exp\{b(\sqrt{1/\al}-1+o(1))\sqrt{\ln n}\}. \end{equation} Letting now $b=1$ if $\al\in(0,1)$ and $b=-1$ if $\al\in(1,2)$, we see that for all large enough $n$ \begin{equation} \frac{\ell(a_n)}{\ell(n)}>\ln n \end{equation} and hence \begin{equation} \sum_n P(|X|>a_n)=\infty, \end{equation} whereas $\int_1^\infty \frac{dt}{t f(t)}<\infty$, so that the inequality \begin{equation} \sum_{n=1}^\infty P\big( |X| > a_n \big) \le \tilde C \int_1^\infty \frac{dt}{t f(t)} \end{equation} cannot hold for any real $\tilde C$.

In the case $\al=1$, instead of \eqref{1} similarly consider \begin{equation} \ell(x)=\exp\frac{\ln x}{\ln\ln x} \end{equation} for $x>e$.