Note that if $x=(x_n)\in\ell^1$, then $x\in\ell^\infty$ as well. Now
$$\|T_nx-x\|_{\ell^\infty}=\|(0,0,,\dots,0,x_n-x_{n+1},x_n-x_{n+2},x_n-x_{n+3},\dots)\|_\infty=\sup_{k\geq1}|x_n-x_{n+k}|$$
But since $x\in\ell^1$ we have that $\{x_n\}$ is a Cauchy sequence (since it converges (to $0$)). So if $\varepsilon>0$ there exists $N\geq1$ so that $|x_n-x_m|<\varepsilon$. Then, for $n\geq N$ we have that
$$\|T_nx-x\|_\infty=\sup_{k\geq1}|x_n-x_{n+k}|\leq\varepsilon$$
which shows that $T_nx\to x$ in $\ell^\infty$. In other words, since this is true for all $x$, we have that $T_n\to I$ strongly, where $I:\ell^\infty\to\ell^\infty$ is the identity operator.
comment: Note that the same would be true if we "enlarged" the domain of each $T_n$ and used the space $c$ of convergent sequences with supremum norm.
Edit
Here are some details about checking whether $T_n$ converges in norm, as OP seems to have trouble with this part.
First, if $T_n$ converges in norm to something, it has to be $I$, because if $T_n$ converges in norm to $S$, then $T_n$ converges strongly to $S$ and therefore $S=I$ (strong operator limits are unique).
First let's compute the norm of $T_n$, but this is not needed to check whether $T_n\to I$ in norm; we should be computing $\|T_n-I\|$ for this, but let's start with that. Note that $n$ is fixed now! We have that $$\|T_n\|=\sup_{\|x\|_{\ell^1}=1}\|T_nx\|_{\ell^\infty}=\sup_{\|x\|=1}\|(x_1,\dots,x_n,x_n,x_n,\dots)\|_{\ell^\infty}=\sup_{\|x\|=1}\bigg\{\max_{1\leq j\leq n}|x_j|\bigg\}\leq\sup_{\|x\|=1}\bigg\{\sum_{j=1}^\infty|x_j|\bigg\}=1 $$
So $\|T_n\|\leq1$. On the other hand, if $x=(1,0,0,\dots)$ then $\|x\|_{\ell^1}=1$ and $T_nx=(1,0,0,\dots,)$ and $\|T_nx\|_{\ell^\infty}=1$, so
$$\|T_n\|=\sup_{\|y\|_{\ell^1}}\|T_n(y)\|\geq\|T_n(x)\|=1$$
and this shows that $\|T_n\|=1$ for all $n$.
Now let's compute in the same fashion the norm of $T_n-I$, again $n$ is fixed. We do not really need to do an exact calculation, we only need to know if $\|T_n-I\|$ converges to $0$ or not. By our earlier computations before the edit,
$$\|T_n-I\|=\sup_{\|x\|=1}\|T_nx-x\|=\sup_{\|x\|=1}\sup_{k\geq1}|x_n-x_{n+k}|$$
Now consider $x=(0,\dots,0,0,1,0,\dots)$ where $1$ appears in the $n+1$ position. Then, $\sup_{k\geq1}|x_n-x_{n+k}|=sup_{k\geq1}|x_{n+k}|=|x_{n+1}|=1$. Also note that $\|x\|_{\ell^1}=1$, so,
$$\|T_n-I\|\geq\|T_nx-x\|_{\ell^\infty}=1$$
This shows that the sequence $\{\|T_n-I\|\}_{n=1}^\infty$ is ALWAYS greater than or equal to $1$, so it cannot converge to $0$.
Best Answer
Fix $\varepsilon>0$. Since $a_n\to0$, there exists $n_0$ such that $|a_n|<\varepsilon^2$ whenever $n>n_0$. So, for $n>n_0$, $$ \|(T-T_n)x\|^2=\sum_{j=n+1}^\infty|a_jx_j|^2\leq \varepsilon^2\,\sum_{j=n+1}^\infty|x_j|^2 \leq\varepsilon^2\,\|x\|^2. $$ This shows that $\|T-T_n\|<\varepsilon$ whenver $n>n_0$, so $\lim T_n=T$.