A standard way to solve this is to consider simultaneously the mean number of rolls $t_n$ needed to reach $n$ for every nonnegative integer $n$ (and to remember at the end that the OP is asking for $t_{30}$). So, let us do that...
For every $n\leqslant0$, $t_n=0$. For every $n\geqslant1$, considering the result of the first roll, one sees that
$$t_n=1+\frac16\sum_{k=1}^6t_{n-k}
$$
Thus, the series $$T(s)=\sum_nt_ns^n$$ solves
$$T(s)=\frac s{1-s}+\frac16\sum_{k=1}^6s^kT(s)$$ from which it follows that
$$T(s)=\frac{6s}{(1-s)\left(6-\sum\limits_{k=1}^6s^k\right)}=\frac{6s}{6-7s+s^7}$$
To extract the coefficient $t_{30}=[s^{30}]T(s)$ of $s^{30}$ in $T(s)$, rewrite this as
$$T(s)=s\left(1-rs\left(1-\frac t7\right)\right)^{-1}=\sum_{n=0}^\infty r^ns^{n+1}\left(1-\frac t7\right)^n$$
where $$r=\frac76\qquad t=s^6$$
hence
$$[s^{30}]T(s)=\sum_{k=0}^4r^{5+6k}[t^{4-k}]\left(1-\frac t7\right)^{5+6k}$$
or, equivalently,
$$[s^{30}]T(s)=\sum_{k=0}^46^{-5-6k}[t^{4-k}]\left(7-t\right)^{5+6k}=\frac7{6^5}\sum_{k=0}^4(-1)^kr^{6k}{5+6k\choose4-k}$$
that is,
$$t_{30}=\frac7{6^5}\left(5-165r^6+136r^{12}-23r^{18}+r^{24}\right)
$$
which can be "simplified" into the exact result $$t_{30}=
\frac{333366007330230566034343}{36845653286788892983296}$$ with a numerical approximation $$t_{30}\approx
9.047634594384022902065997942672796588425278684184104625$$
Edit: To get some estimates of $t_n$ when $n\to\infty$, note that $$T(s)=\frac s{(1-s)^2Q(s)}$$ where $$Q(s)=\frac16(6+5s+4s^2+3s^3+2s^4+s^5)$$ Furthermore, $(1-s)Q(s)=1-\frac16\sum\limits_{k=1}^6s^k$ has no zero in the closed unit disk except the simple zero $s=1$ hence $Q(s)$ has no zero in the closed unit disk. This implies that $$T(s)=\frac a{(1-s)^2}+\frac b{1-s}+R(s)$$ for some given $(a,b)$ and some rational fraction $R(s)=\sum\limits_nr_ns^n$ with no pole in the closed unit disk. Then, there exists some finite $c$ and some $\varrho$ in $(0,1)$ such that, for every $n$, $$|r_n|\leqslant c\varrho^n$$ This yields, again for every $n$, $$|t_n-a(n+1)-b|=|r_n|\leqslant c\varrho^n$$ Equivalently, $$t_n=an+(a+b)+O(\varrho^n)$$ To identify $(a,b)$, note that $$(1-s)^2T(s)=a+b(1-s)+(1-s)^2R(s)$$ hence $$a=\left.(1-s)^2T(s)\right|_{s=1}=\frac1{Q(1)}$$ and $$b=-\left.\frac d{ds}[(1-s)^2T(s)]\right|_{s=1}=-\frac1{Q(1)}+\frac{Q'(1)}{Q(1)^2}$$ or, equivalently, $$a+b=\frac{Q'(1)}{Q(1)^2}$$
Finally, $Q(1)=\frac72$ and $Q'(1)=\frac{35}6$ hence $a=\frac27$ and $a+b=\frac{10}{21}$, which implies
$$t_n=\frac27n+\frac{10}{21}+O(\varrho^n)$$
Edit-edit: More generally, throwing repeatedly a "die" producing a random number of points distributed like $X$, following the same route, one gets
$a=E(X)$ and $a+b=Q'(1)/E(X)^2$ with $Q'(1)=\frac12E(X(X-1))$, hence, for some $\varrho_X$ in $(0,1)$ depending on the distribution of $X$, $$t_n=\frac1{E(X)}n+\frac{E(X(X-1))}{2E(X)^2}+O(\varrho_X^n)$$
