Let $p_n = \frac{1}{6}\left(p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6} \right)$.
Let $p_0 = 1$ and $p_k = 0$ for $k < 0$.
Using this recursive equation, is there a simple way to obtain the following expressions, in particular for $p_3$ and onward?
$$p_1 = \frac{1}{6}, \quad p_2 = \frac{1}{6}\left(1 + \frac{1}{6}\right), \quad p_3 = \frac{1}{6}\left(1 + \frac{1}{6}\right)^2$$
$$p_4 = \frac{1}{6}\left(1 + \frac{1}{6}\right)^3, \quad p_5 = \frac{1}{6}\left(1 + \frac{1}{6}\right)^4, \quad p_6 = \frac{1}{6}\left(1 + \frac{1}{6}\right)^5$$
$$p_7 = \frac{1}{6}\left(\left(1 + \frac{1}{6}\right)^6 - 1 \right)$$
I'm able to derive these directly via algebraic manipulations of the equation for $p_n$, but this gets messy beyond $p_3$ so I'm hoping someone can describe a more elegant approach.
Computing, we have (for sure) $$p_n=\frac {a_n}{6^n}$$ and the $a_n$ correspond to the sequence $$\{1,1,7,49,343,2401,16807,70993,450295,2825473,17492167\}$$ which is not identified by $OEIS$.
Coefficients $a_n$ seem to vary exponentially but I did not find any way to make the realtion explicit.
In any manner, the solution of $$p_n = \frac{1}{6}\left(p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6} \right)$$ is given by $$p_n=c_0+\sum_{i=1}^5 c_i \,r_i^n$$ where the $r_i$ are the roots of the quintic polynomial $$1+2x+3x^2+4x^3+5x^4+6x^5=0$$ One of them is real and four are complex conjugate.
The best I have able to do is to compute the $r_i$'s and to make them rational to get $$r_1\sim-\frac{2316}{3455}\qquad r_{2,3}\sim -\frac{405}{1078}\pm\frac{65 }{114}i\qquad r_{4,5}\sim \frac{223}{758}\pm\frac{131 }{196}i$$
Edit
Computing the $a_n$ up to $n=30$ and performing a quick and dirty linear regression for $$\log(a_n)=\alpha + \beta n$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & -1.29097167 & 0.01606328 & \{-1.32437708,-1.25756626\} \\ \beta & +1.79344429 & 0.00081322 & \{+1.79175311,+1.79513548\} \\ \end{array}$$ and $\beta$ is "clearly" $\log(6)$. Fixing $\beta$ at this value leads to $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & -1.25980243 & 0.00617784 & \{-1.27264996,-1.24695490\} \end{array}$$ and $e^\alpha$ is very close to $\frac 2 7$ as given by @Ross Millikan.
Update
In fact, if we consider the shorter sequences $$p_n^{(k)}=\frac 1 k \sum_{i=1}^{k}p_{n-i}^{(k)}$$ with $$p_0^{(k)}=1 \quad p_1^{(k)}=\frac 1 k\quad p_2^{(k)}=\frac{1}{k}\left(1+\frac{1}{k}\right)\quad p_3^{(k)}=\frac{1}{k}\left(1+\frac{1}{k}\right)^2\quad \cdots$$ for which the general term can be computed, $$\color{blue}{\lim_{n\to \infty } \,p_n^{(k)}=\frac 2 {k+1}}$$