The dimesion is indeed one.

One can show that the element \begin{align*} \alpha_1 &= 1,\\ \alpha_n &= (1-n)\int_0^1 \frac{t^{n-1}}{\log(1-t)} dt\quad\text{for } n = 2,3,\dots \end{align*} lies in $E$. So let $x\in E$ with $x_1=0$. We need to show that $x=0$. Let $y_k=\frac {k+1}{k}x_{k+1}$. Then $y\in\ell^\infty$ and $Tx=x$ as well as $x_1=0$ lead to \begin{align*} 0&=\sum_{k=1}^\infty\frac{y_{k}}{k(k+1)},\\ \frac m{m+1}y_m&=\sum_{k=m+1}^\infty\frac{y_k}{(k-m)(k-m+1)},\quad m\ge 1. \end{align*} We have to show $y=0$. For $m\in{\mathbb N}_0$ let $z^{(m)}\in\ell^1$ be defined by \begin{align*} z^{(0)}_k&=\frac1{k(k+1)},\\ z^{(m)}_k&=\begin{cases}0&k<m,\\ \frac m{m+1}&k=m,\\ \frac{-1}{(k-m)(k-m+1)}&k>m. \end{cases} \end{align*} Now $\ell^\infty$ is the dual space of $\ell^1$. Let $\langle .,.\rangle$ denote the duality pairing. The formulas above say that $$ \langle{z^{(k)},y}=\rangle0\qquad\text{for }k=0,1,\dots $$ Therefore the theorem follows if we can show that the sequence $z^{(0)},z^{(1)},\dots$ spans a dense subspace of $\ell^1$. Let $Z$ denote the span of $z^{(0)},z^{(1)},\dots$. We will now show that the first canonical basis element $e_1=(1,0,\dots)$ lies in the closure of $Z$. We have $$ \begin{array}{ccccccccc} z^{(1)}&=&\frac12,&\frac{-1}2,&\frac{-1}{6},&\frac{-1}{12},&\frac{-1}{20},&\frac{-1}{30},&\dots\\ &\\ z^{(2)}&=&0,&\frac23,&\frac{-1}2,&\frac{-1}{6},&\frac{-1}{12},&\frac{-1}{20},&\dots\\ &\\ z^{(3)}&=&0,&0,&\frac34,&\frac{-1}2,&\frac{-1}{6},&\frac{-1}{12},&\dots\\ &\\ z^{(4)}&=&0,&0,&0,&\frac45,&\frac{-1}2,&\frac{-1}{6},&\dots\\ &\\ z^{(5)}&=&0,&0,&0,&0,&\frac56,&\frac{-1}2,&\dots\\ \end{array} $$ Let $a^{(1)},a^{(2)},\dots\in\ell^1$ be defined by $a^{(1)}=z^{(1)}$ and for $m\ge 2$ inductively by \begin{align*} a^{(m)}_k=\begin{cases} \frac12&k=1,\\ 0&2\le k\le m,\\ a^{(m-1)}_k+a_m^{(m-1)}\frac{m+1}m\frac1{(k-m)(k+1-m)}& k\ge m+1. \end{cases} \end{align*} Then $a^{(m)}=a^{(m-1)}-a_m^{(m-1)}\frac{m+1}mz^{(m)}$, so that $a^{(m)}\in Z$. Iterating the recursive relation, we get $$ a^{(m)}=a^{(1)}-a_2^{(1)}\frac32 z^{(2)}-\dots-a_m^{(m-1)}\frac{m+1}m z^{(m)}. $$ Setting $b_j=a_{j+1}^{(j)}$ this becomes $$ a^{(m)}=z^{(1)}-\sum_{j=1}^{m-1}b_j\frac{j+2}{j+1}z^{(j+1)}. $$ If we can show that the sequence $a^{(m)}$ converges in $\ell^1$, it follows that it converges to $\frac12e_1$, where $e_1=(1,0,0,\dots)$ is the first standard basis vector, so that we have $$ e_1\in\overline Z. $$ As the $\ell^1$-norms of ths $z^{(j)}$ are globally bounded, in order to show convergence of $(a^{(m)})$, it suffices to show $\sum_{j=1}^\infty|b_j|<\infty$. For this we read the recursive relation for $k>m$ as $$ a^{(m)}_k=\frac{-1}{k(k-1)}+\sum_{j=1}^{m-1}b_j\frac{j+2}{j+1}\frac1{(k-j-1)(k-j)}. $$ In particular, for $k=m+1$ we get $$ b_m=\frac{-1}{m(m+1)}+\sum_{j=1}^{m-1}b_j\frac{j+2}{j+1}\frac1{(m-j)(m+1-j)}. $$ It follows $b_j<0$ for every $j$ and with $c_j=|b+j|$ we have \begin{align*} c_m&=\frac1{m(m+1)}+\sum_{j=1}^{m-1}c_j\frac{j+2}{j+1}\frac1{(m-j)(m+1-j)}\\ &=\frac1{m(m+1)}+\sum_{j=1}^{m-1}c_j\frac{j+2}{j+1}\frac1{(m+1-j)(m+2-j)}\underbrace{\frac{m+2-j}{m-j}}_{\le\frac32}\\ &\le\frac1{m(m+1)}+\frac32\sum_{j=1}^{m-1}c_j\frac{j+2}{j+1}\frac1{(m+1-j)(m+2-j)}. \end{align*} Using $\frac1{m(m+1)}=\frac1m-\frac1{m+1}$ and collapsing the ensuing telescope sums we get \begin{align*} \sum_{m=1}^Nc_m&\le\sum_{m=1}^N\frac1{m(m+1)} +\frac32\sum_{j=1}^{N-1}c_j\frac{j+2}{j+1}\sum_{m=j+1}^N\frac1{(m+1-j)(m+2-j)}\\ &=\left(1-\frac1{N+1}\right)+\frac32\sum_{j=1}^{N-1}c_j\frac{j+2}{j+1} \left(\frac12-\frac1{N+2-j}\right)\\ &\le 1+\frac34\sum_{j=1}^{N-1}c_j\left(1+\frac{1}{j+1}\right). \end{align*} Subtracting $\frac34\sum_{j=1}^{N-1}c_j$ and multiplying by 4 we get \begin{align*} \sum_{m=1}^Nc_m&\le \sum_{m=1}^Nc_m+3c_N\\ &\le 4+3\sum_{j=1}^{N-1}\frac{c_j}{j+1}\\ &=4+\frac32c_1+c_2+\frac34c_3+\dots+\frac3{N}c_{N-1}\\ &\le4+\frac34c_1+\frac14c_2+\frac34\sum_{j=1}^{N}c_j. \end{align*} Subtracting $\frac34\sum_{j=1}^{N-1}c_j$ and multiplying by 4 we get \begin{align*} \sum_{m=1}^Nc_m&\le 16+3c_1+c_2. \end{align*} Since this holds for every $N$, we conclude $\sum_{m=1}^\infty c_m<\infty$. So we have shown $e_1\in\overline Z$. The same argument mutatis mutandis shows $ej\in\overline Z$ for all $j$. The claim follows.

Theorem. Let $T:\ell^\infty\to\ell^\infty$ be defined by $$ (Tx)_m=\sum_{k=m+1}^\infty p_{k,m}x_k, $$ where $p_{k,m}=\frac k{(k-1)(k-m)(k-m+1)}$. Let $L=\{x\in\ell^\infty: Tx=x\}$. Then $\dim L= 1$.

We first show $\dim L\ge 1$.

Lemma. The element $\alpha$ with $\alpha_1 = 1$ and \begin{align*} \alpha_k &= (1-k)\int_0^1 \frac{t^{k-1}}{\log(1-t)} dt\quad\text{for } k = 2,3,\dots \end{align*} lies in $L$.

For $m=1$ we compute \begin{align*} (T\alpha)_1&=\sum_{k=2}^\infty \frac 1{(k-1)^2}\alpha_k\\ &=-\sum_{k=2}^\infty \frac1{k-1}\int_0^1\frac{t^{k-1}}{\log(1-t)}\,dt\\ &=-\int_0^1\sum_{k=2}^\infty \frac{t^{k-1}}{k-1}\frac{1}{\log(1-t)}\,dt=\int_0^11\,dt=1=\alpha_1. \end{align*} For $N\in{\mathbb N}$, $N\ge 2$ and $x\in\ell^\infty$ we get \begin{align*} \sum_{m=1}^N(Tx)_m&=\sum_{m=1}^N\sum_{k={m+1}}^\infty\frac k{(k-1)(k-m)(k+1-m)}x_k\\ &=\sum_{k=2}^\infty\frac{kx_k}{k-1}\sum_{m=1}^{\min(N,k-1)}\frac1{(k-m)(k+1-m)}\\ &=\sum_{k=2}^\infty\frac{kx_k}{k-1}\sum_{m=1}^{\min(N,k-1)}\left(\frac1{k-m}-\frac1{k+1-m}\right)\\ &=\sum_{k=2}^\infty\frac{kx_k}{k-1}\left(\frac1{k-\min(N,k-1)}-\frac1k\right)\\ &=\sum_{k=2}^N\frac{kx_k}{k-1}\left(1-\frac1k\right)+\sum_{k=N+1}^\infty\frac{kx_k}{k-1}\left(\frac1{k-N}-\frac1k\right)\\ &=\sum_{k=2}^Nx_k+N\sum_{k=N+1}^\infty\frac{x_k}{(k-1)(k-N)}. \end{align*} Now assume we have shown $(T\alpha)_j=\alpha_j$ for $1\le j\le N-1$. Then we get $$ 1+(T\alpha)_N=\alpha_N+N\sum_{k=N+1}^\infty\frac{\alpha_k}{(k-1)(k-N)}. $$ We compute \begin{align*} N\sum_{k=N+1}^\infty\frac{\alpha_k}{(k-1)(k-N)} &=-N\sum_{k=N+1}^\infty\frac{1}{(k-N)} \int_0^1 \frac{t^{k-1}}{\log(1-t)} dt\\ &=-N\int_0^1\sum_{k=N+1}^\infty\frac{t^{k-N}}{(k-N)} \frac{t^{N-1}}{\log(1-t)} dt\\ &= N\int_0^1t^{N-1}\,dt=1. \end{align*} The lemma follows. $\square$

In order to show $\dim L\le 1$, let now $x\in L$ with $x_1=0$. We need to show that $x=0$. Let $y_k=\frac {k+1}{k}x_{k+1}$. Then $y\in\ell^\infty$ and $Tx=x$ as well as $x_1=0$ lead to \begin{align*} 0&=\sum_{k=1}^\infty\frac{y_{k}}{k(k+1)},\\ \frac m{m+1}y_m&=\sum_{k=m+1}^\infty\frac{y_k}{(k-m)(k-m+1)},\quad m\ge 1. \end{align*} We have to show $y=0$. For $m\in{\mathbb N}_0$ let $z^{(m)}\in\ell^1$ be defined by \begin{align*} z^{(0)}_k&=\frac1{k(k+1)},\\ z^{(m)}_k&=\begin{cases}0&k<m,\\ \frac m{m+1}&k=m,\\ \frac{-1}{(k-m)(k-m+1)}&k>m. \end{cases} \end{align*} Now $\ell^\infty$ is the dual space of $\ell^1$. Let $<.,.>$ denote the duality pairing. The formulas above say that $$ <z^{(k)},y>=0\qquad\text{for }k=0,1,\dots $$ Therefore the theorem follows if we can show that the sequence $z^{(0)},z^{(1)},\dots$ spans a dense subspace of $\ell^1$. Let $Z$ denote the span of $z^{(0)},z^{(1)},\dots$.

For $m\in{\mathbb N}$ let $$ h^{(m)}_k=\begin{cases}0&1\le k\le m-1,\\ \frac m{(k-m+1)(k+1)}&k\ge m. \end{cases} $$ Then $h^{(1)}=z^{(0)}$ and a quick computation shows $h^{(m)}=h^{(m-1)}-z^{(m-1)}$ for $m\ge 2$. Therefore, by induction we have $h^{(m)}\in Z$ for every $m$. Next, let $$ v^{(m)}=\frac1m h^{(m)}+z^{(m)}. $$ then $v^{(m)}\in Z$ for every $m\in{\mathbb N}$ and $$ v^{(m)}_k=\begin{cases}0&1\le k\le m-1,\\ 1& k=m,\\ \frac{-(m+1)}{(k+1)(k-m)(k-m+1)}&k\ge m+1.\end{cases} $$ Let $n\in{\mathbb N}$ and let $e_n$ be the $n$-th standard basis element of $\ell^1$, i.e., $e_n=(0,\dots,0,1,0,\dots)$ with the $1$ at the $n$-th position. We show that $e_n$ lies in the closure of the span of $v^{(n)},v^{(n+1)},\dots$. For this purpose let $a^{(n)}=v^{(n)}$ and for $m>n$ let $a^{(m)}=a^{(m-1)}-a^{(m-1)}_mv^{(m)}$. Then $a^{(m)}\in Z$ and $$ a^{(m)}_k=\begin{cases}0&1\le k\le m,\ k\ne n,\\ 1 &k=n,\\ a^{(m-1)}_k-a_m^{(m-1)}v^{(m)}_k&k\ge m+1.\end{cases} $$ Iterating the recursive relation yields for $k>m$, \begin{align*} a^{(m)}_k&=a^{(n)}_k-\sum_{j=n}^{m-1}a^{(j)}_{j+1}v_k^{(j+1)}\\ &=\frac{-(n+1)}{(k+1)(k-n)(k-n+1)} +\sum_{j=n}^{m-1}a_{j+1}^{(j)}\frac{j+2}{(k+1)(k-j-1)(k-j)}. \end{align*} For $m\ge n$ set $b_m=a^{(m)}_{m+1}$, then in particular, $$ b_m=\frac{-(n+1)}{(m+2)(m-n+1)(m-n+2)}+\sum_{j=n}^{m-1}\frac{b_j(j+2)}{(m+2)(m-j)(m-j+1)}. $$ Lemma. For every $m\ge n$ we have $|b_m|\le \frac{n+1}{m+2}$.

Proof of the lemma. For $m=n$ we have \begin{align*} |b_m|=\frac{n+1}{n+2}\frac12\le \frac{n+1}{n+2}=\frac{n+1}{m+2}. \end{align*} For the induction step we assume $m>n$ and the claim proven for smaller indices. Then \begin{align*} |b_m|&\le \frac{n+1}{(m+2)(m-n+1)(m-n+2)}+\sum_{j=n}^{m-1}\frac{|b_j|(j+2)}{(m+2)(m-j)(m-j+1)}\\ &\le \frac{n+1}{m+2}\left(\frac1{m-n+1}+\sum_{j=n}^{m-1}\frac1{m-j)(m-j+1)}\right)\\ &= \frac{n+1}{m+2}\left(\frac1{m-n+1}+1-\frac1{m-n+1}\right)=\frac{n+1}{m+2}. \end{align*} $\square$

One has \begin{align*} \parallel{e_n-a^{(m)}}\parallel_1&=\sum_{k=m+1}^\infty |a_k^{(m)}|\\ &\le \sum_{k=m+1}^\infty\frac{n+1}{(k+1)(k-n)(k+1-n)}\\ &\ \ \ +\sum_{k=m+1}^\infty \sum_{j=n}^{m-1}|b_j| \frac{j+2}{(k+1)(k-j-1)(k-j)}. \end{align*} We estimate the first sum \begin{align*} \sum_{k=m+1}^\infty\frac{n+1}{(k+1)(k-n)(k+1-n)} &\le \frac{n+1}{m+2}\sum_{k=m+1}^\infty \frac1{(k-n)(k+1-n)}\\ &= \frac{n+1}{m+2}\frac1{m+1-n}=A(m,n)\to 0,\quad m\to\infty. \end{align*} For the second sum we have \begin{align*} \sum_{k=m+1}^\infty \sum_{j=n}^{m-1}|b_j| \frac{j+2}{(k+1)(k-j-1)(k-j)} &\le \sum_{k=m+1}^\infty \sum_{j=n}^{m-1} \frac{n+1}{(k+1)(k-j-1)(k-j)}\\ &=\sum_{k=m+1}^\infty \frac{n+1}{k+1}\left(\frac1{k-m}-\frac1{k-n}\right), \end{align*} which tends to zero as $m\to\infty$. We have shown that $$ \parallel{e_n-a^{(m)}}\parallel_1\to 0 $$ as $m\to\infty$. The theorem follows. $\square$