If $p>1-5d$, then $a_n(p)=0$ for all $n\ge1$ and hence $E_n(p)=0$ for all $n\ge1$.
So, without loss of generality $0\le p\le1-5d$. Then, letting $E_{-1}:=1$, for all $n\ge0$ we get \begin{equation*} E_n=\sum_{i=-1}^{n-1}l_{n-1-i}E_i, \end{equation*} where $E_n:=E_n(p)$ and $l_n:=l_n(p)$.
Here is the key point: Using the fact that $(l_5,l_6,\dots)$ is a geometric progression, one can check that \begin{equation*} E_{n+1}=\sum_{j=0}^5 b_j E_{n-j} \tag{1} \end{equation*} for all $n\ge4$, where \begin{equation*} b_j:=l_j+(p+5d-1)l_{j-1}. \end{equation*} Moreover, one can check that \begin{equation} \text{$b_0,\dots,b_5$ are $>0$ and $\sum_{j=0}^5 b_j=1$. }\tag{2} \end{equation}
The general solution of the linear difference equation (1) is given by \begin{equation*} E_n=\sum_{k=0}^5 C_k(n) z_k^n, \tag{3} \end{equation*} where the $z_k$'s are the roots of the characteristic polynomial \begin{equation*} P(z):=z^6-\sum_{j=0}^5 b_j z^{5-j}=z^6-\sum_{j=0}^5 b_{5-j}z^j \end{equation*} and the $C_k(n)$'s are polynomials in $n$ (whose degrees are nonzero if the corresponding roots $z_k$ are multiple ones). Moreover, in view of (2), we have the following:
(i) $1$ is a root of $P(z)$; say, $z_0=1$.
(ii) The multiplicity of $z_0=1$ is $1$, since $P'(1)=6-\sum_{j=0}^5 b_j(5-j)\ge6-5\sum_{j=0}^5 b_j=1>0$. So, $C_0:=C_0(n)$ does not depend on $n$.
(iii) $|z_k|<1$ for all $k=1,\dots,5$: Indeed, if $0=P(z)=z^6-\sum_{j=0}^5 b_{5-j}z^j$ for some complex $z$ with $|z|\ge1$, then, by the triangle inequality, $|z|^6\le\sum_{j=0}^5 b_{5-j}|z|^j\le\sum_{j=0}^5 b_{5-j}|z|^5=|z|^5$, so that $|z|\le1$ and hence $|z|=1$. Moreover, similarly the condition $|z|=1$ for a root $z$ of $P(z)$ implies the contradictory inequality $|z|^6<|z|^5$ unless the complex numbers $1$ and $z$ lie on the same ray emanating from $0$ -- that is, unless $z=1$.
It follows from (3) and (i)--(iii) that \begin{equation} E_n\to C_0 \tag{4} \end{equation} as $n\to\infty$. Of course, $E_n$ and $C_0$ may/will depend on $d,p$. The convergence in (4) holds for each pair $(d,p)$ satisfying the OP conditions $0<d<1/5$ and $0\le p\le1-5d$.