For $n\leq 6$, clearly $P(1) = 0$ and $$P(2) = \frac16\\ P(3) = \frac16+\frac1{36} = \frac7{36}\\ \vdots\\ P(6) = \frac16+\frac1{36}+\frac{1}{216}+\frac{1}{1296}+\frac{1}{7776} $$ That is, for $2\leq n \leq 6$ $$ P(n) = \frac16 \sum_{k=0}^{n-2}\frac1{6^k} = \frac15 \left( 1-\left(\frac16\right)^{n-1}\right) $$
For $n>6$ the only routes to a score of $n$ are starting with between $n-6$ and $n-2$ ones, and then rolling the right number to land at $n$. This gives $$ P(n) =\frac16 \left[ \left(\frac16\right)^{n-6}+\left(\frac16\right)^{n-5}+\left(\frac16\right)^{n-4}+\left(\frac16\right)^{n-3}+\left(\frac16\right)^{n-2} \right]= \frac{1555}{6^{n-1}} $$
So the full distribution is $$ P(n) = \left\{ \begin{array}{cl} 0 & \mbox{if } n<2 \\ \frac15 \left( 1-\left(\frac16\right)^{n-1}\right) & \mbox{if } 2\leq n \leq 6 \\ \frac{1555}{6^{n-1}} & \mbox{if } n > 6 \end{array} \right. $$
** EDIT NOTE ** IN response to a correction by O. Von Seckendorff, I fixed the exponent in the $n>6$ case, which had erroneously been $n+3$ instead of $n-1$.
