Let $c(x)$ be the number of sequences that sum to $x$. How precisely do you need to know $c(x)$? These things tend to grow exponentially, so we should expect $c(10^{18})$ to have about a billion billion digits, up to some smallish factor. An exact answer will be practically impossible to compute. But you can possibly get at least the exact number of digits and the first few significant digits by this method:
Since the order of the term matters to you, the ${1,2}$ case is easy to solve: $c(x)$ is simply the $x+1$th Fibonacci number. You can see this because the last term in a sum totalling $x$ must be either $1$ or $2$, so therefore $c(x)=c(x-1)+c(x-2)$, which is the recurrence formula for the Fibonacci numbers. This agrees with your brute-force solution for $c(4)$, because $F_5=5$.
This suggests a general plan of writing down the linear recurrence that arises from your set of terms. In general the solutions of such a recurrence are linear combinations of exponential functions, whose bases are the roots of the characteristic polynomial. For large $x$ such as $10^{18}$, the term with the largest base will dominate all others completely, so we can just ignore all of the others. The trick is then just to find its coefficient.
One way to do this would be to set $c(x)=0$ for $-15\le x<0$ and $c(0)=0$, and then solve 16 linear equations in 16 unknowns.
Another way is simply to run the recurrence up to $c(N)$ for some $N$ between, say, 100 and 1000, and assume that by that time all of the smaller bases have faded into insignificance. Then we can estimate the coefficient as $\frac{c(N)}{r^N}$, and our approximation for large $x$ is $\log c(x) \approx \log c(N) +(x-N)\log r$ It is necessary to work with logarithms in this step, because even $c(1000)$ is likely to overflow the hardware floating-point representation in your computer. In fact, one could choose $N$ adaptively such that $c(N)$ is around $2^{500}$.
You can use $\frac{c(N)}{c(N-1)}$ as a first approximation of $r$ and refine it using Newton-Raphson iteration on the characteristic polynomial.
Special cases: If the numerically largest root of the characteristic polynomial is complex, the recurrence can have oscillating solutions that are never dwarfed by the exponential term. I think the only way this can happen is if the $a_i$s share a divisor $\ne 1$. So start by dividing out the greatest common divisor from the $a_i$s as well as from $x$.
If the largest root of the characteristic polynomial is a multiple root, the growth rates from the master theorem look slightly different. I think this cannot happen when the recurrence is constructed from your problem, but it is probably worth checking for explicitly when you have found $r$.