What are the odds of hitting exactly 100 rolling a fair die

The probability of hitting $n$ on the nose is a function of $n$, call it $f(n)$. It satisfies the recurrence $$f(n)=\frac{f(n-1)+f(n-2)+f(n-3)+f(n-4)+f(n-5)+f(n-6)}6$$ with boundary values $f(0)=1$ and $f(n)=0$ for $n\lt0$. I didn't try to solve the recurrence, but my numerical calculations agree with yours, it looks like $f(n)\to2/7$.

The generating function $$H(x)=\sum\limits_{n=0}^\infty h_nx^n,$$ where $h_0=1$ and, for every $n\geqslant1$, $h_n$ is the probability to hit exactly $n$, solves the identity $$H(x)=1+P(x)H(x),\qquad P(x)=\frac16(x+x^2+\cdots+x^6),$$ hence $$H(x)=\frac1{1-P(x)}.$$ The limit of $h_n$ when $n\to\infty$ is $$\ell=\lim_{x\to1}(1-x)H(x)=\frac1{P'(1)},$$ that is, $$\ell=\frac6{1+2+\cdots+6}=\frac27.$$ This extends to every "die" producing any collection of numbers in any biased or unbiased way. If the "die" produces a random positive integer number $\xi$, the limit of $h_n$ becomes $$\ell=\frac1{E(\xi)}.$$ Assuming the limit $\ell$ exists, this can be understood intuitively as follows: by the law of large numbers, the sum of $n$ draws is about $k=nE(\xi)$ hence, after $n$ draws, $n$ large, one has hit $n$ values from roughly $k$. If each value has a chance roughly $\ell$ to be hit, one can expect that $\ell\approx n/k$, QED. Obvious counterexamples are when $\xi$ is, say, always a multiple of $3$, and these are essentially the only cases since $\ell$ exists if and only if the greatest common divisor of the support of $\xi$ is $1$.

Edit: To estimate the difference $h_n-\ell$ in the usual case, note that $$1-P(x)=(1-x)(1+x/a)(1-x/u)(1-x/\bar u)(1-x/v)(1-x/\bar v),$$ for some $a$ real positive and some complex numbers $u$ and $v$ with nonzero imaginary parts, hence $$H(x)=\frac{\ell}{1-x}+\frac{b}{1+x/a}+\frac{r}{1-x/u}+\frac{\bar r}{1-x/\bar u}+\frac{s}{1-x/v}+\frac{\bar s}{1-x/\bar v},$$ for some real number $b$ and some complex numbers $r$ and $s$ defined as $$b=-\frac1{aP'(-a)},\qquad r=\frac1{uP'(u)},\qquad s=\frac1{vP'(v)}.$$ Thus, for every $n$, $$h_n=\ell+b(-1)^na^{-n}+2\Re\left(r u^{-n}+s v^{-n}\right).$$ Numerically, $a\approx1.49$, and $|u|$ and $|v|$ are approximately $1.46$ and $1.37$, hence all this yields finally $$|h_n-\ell|=O(\kappa^{-n}),\qquad\kappa\approx1.37.$$ For $n=100$, $\kappa^{-n}\approx2\cdot10^{-14}$ hence one expects that $h_n$ coincides with $\ell$ at up to $13$ or $14$ decimal places.

Some addition on the answer of @bof:

For $n\in\mathbb{Z}$ let $E_{n}$ denote the event that number $n$ shows up in the sequence.

This with $E_{n}=\emptyset$ if $n<0$ and $E_{0}=\Omega$ (we start the list with $0$).

Let $X$ denote the first number that appears.

Then $P\left(E_{n}\right)=\sum_{i=1}^{6}P\left(E_{n}\mid X=i\right)P\left(X=i\right)=\frac{1}{6}\sum_{i=1}^{6}P\left(E_{n-i}\right)$