It is obvious that $\Vert Tx\Vert_\infty\leq\Vert x\Vert_\infty$, hence
$\Vert T\Vert\leq 1$. Since $\Vert T(1,0,0,\ldots)\Vert_\infty=\Vert (1,0,\ldots)\Vert_\infty$, we conclude that $\Vert T\Vert=1$.
One can easily verify that
$$T(l^\infty)=\left\{(\xi_j):\exists C>0 \text{ such that } |\xi_j|\leq C/j\text{ for all j}\right\}$$
$T(l^\infty)$ is not closed in $l^\infty$: Let $x_n=(1,\dfrac{1}{\sqrt{2}},\ldots,\dfrac{1}{\sqrt{n}},0,0,\ldots)=T(1,\sqrt{2},\ldots,\sqrt{n},0,\ldots)$. The sequence $(x_n)$ obviously converges to $x=\left(1/\sqrt{j}\right)_{j=1}^\infty\in l^\infty$, which does not lie in $T(l^\infty)$ (if it did, what would be it's pre-image?).
The inverse operator $T^{-1}$ is not bounded: Consider the sequence $(x_n)\subseteq T(l^\infty)$ as above. This sequence is bounded but the image
$\left\{T^{-1}x_n\right\}$ is not, since $\Vert T^{-1}x_n\Vert_\infty=\sqrt{n}$.
You could also argue in this way: If $T^{-1}$ were bounded, then for every Cauchy sequence $(y_n)\in T(l^\infty)$, the sequence $(T^{-1} y_n)$ would also be Cauchy, and hence would converge to some $x\in l^\infty$ since $l^\infty$ is complete. But then, since $T$ is bounded, $y_n$ would converge to $Tx$. Then $T(l^\infty)$ would be complete, contradicting the fact that it is not closed in $l^\infty$ (this argument can actually be used to show that if $T:X\rightarrow Y$ is an isomorphism between a Banach space $X$ and some normed space $Y$ such that $T$ and $T^{-1}$ are bounded, then $Y$ is Banach).
Hint for $\implies$: If $\lambda_n \not \to 0,$ then for some $\epsilon>0,|\lambda_{n_k}| > \epsilon$ along a subsequence $n_k.$ Letting $e_n$ denote the usual "basis" vector, consider the sequence $e_{n_k}$ in the unit ball of $l^p$ and its images under $T.$
Best Answer
In the first comment I suggested the following strategy: write $T=\sum_j T_j$, where $T_j$ is a linear operator defined by $T_jx=\{k_jx_{n-j}\}$. You should check that this is indeed correct, i.e., summing $T_j$ over $j$ indeed gives $T$. Next, show that $\|T_j\|=|k_j|$ using the definition of the operator norm. Finally, use the triangle inequality $\|Tx\|_{\ell^p}\le \sum_j \|T_jx\|_{\ell_p}$.