Let's start by defining $r(n)$ to be the probability that the sum is ever exactly $n$.
Let's express the probability that the sum will ever be $a$, $a+1$, or $a+2$ using $r$. Let's call that probability $P$.
For a set of positive integers $S$, define $p_s$ to be the probability that the sum will ever be in $S$.
Using the inclusion exclusion principal, we have
$$
P = p_{\{a\}}+p_{\{a+1\}}+p_{\{a+2\}}-p_{\{a,a+1\}}-p_{\{a,a+2\}}-p_{\{a+1,a+2\}}+p_{\{a,a+1,a_2\}}.$$
We can calculate all of these probabilities using the function $r$.
We have
$$
p_{\{a\}} = r(a), p_{\{a+1\}} = r(a+1), \text{ and } p_{\{a+2\}} = r(a+2).$$
Further,
$$p_{\{a,a+1\}} = r(a) \frac{1}{6}$$
and
$$p_{\{a+1,a+2\}} = r(a+1) \frac{1}{6}.$$
Since we can get to $a+2$ from $a$ in two ways, we have
$$
p_{\{a,a+2\}} = r(a)\frac{1}{6}+ r(a)\frac{1}{36} = \frac{7}{36}r(a).
$$
Finally,
$$p_{\{a,a+1,a_2\}} = r(a) \frac{1}{36}.$$
Putting this all together, we have
$$
P = \frac{2}{3}r(a)+\frac{5}{6}r(a+1)+r(a+2).
$$
Now, here's the complicated part. The closed form for $r$ is
$$
r(n) = \frac{2}{7} + \frac{1}{7} \sum_{i=2}^{7} A_i^n
$$
where the $A_i$ are complex numbers inside the unit circle. From this, we know that as $n$ goes to infinity, $r(n)$ goes to $\frac{2}{7}$, and so, as $a$ goes to infinity, $P$ approaches $\frac{5}{7}$.
If we really want to calculate $P$ exactly for some specific $a$, we can use the relation $\frac{2}{3}r(a)+\frac{5}{6}r(a+1)+r(a+2)$ given earlier and then go about calculating these $r$ values using the recurrence
$$
r(n) = \frac{1}{6}\sum_{i=1}^6 r(n-i)
$$
for $n>6$
with
$$r(1)= \frac{1}{6}
$$
$$
r(2)=\frac{1}{6}+\frac{1}{6} r(1) = \frac{7}{36}
$$
$$
r(3) = \frac{1}{6} + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{49}{216}
$$
$$
r(4) = \frac{1}{6} + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{343}{1296}
$$
$$
r(5) = \frac{1}{6} + \frac{1}{6} r(4) + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{2401}{7776}
$$
$$
r(6) = \frac{1}{6} + \frac{1}{6} r(5) + \frac{1}{6} r(4) + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1)
= \frac{16807}{46656}
$$
This way, we can calculate, for instance, that the probability of ever having a sum of $4$, $5$ or $6$ is exactly
$$
\frac{343}{432}
$$
and of ever having a sum of $1000$, $1001$ or $1002$ is
approximately $\frac{5}{7}+4.139\times10^{-138}$ (the exact fraction has
too many digits in its numerator and denominator to display here).
