Inspired by the PIE-based answer of @RobPratt I found the following much simpler (computationally speaking) solution!
As usual all s-f's considered here are in spades.
Consider the value $5 \times {52-9 \choose 13-9}$. This is the number of pairs of sets $(X,Y)$, where $|X| = 9$ and the contents form a $9$-long s-f and $|Y| = 4$ and the contents are arbitrary, as long as $X, Y$ are disjoint. The first factor in the product is $5$ because there are $5$ possible $9$-long s-f's.
Now, every hand with an exactly $9$-long s-f can be represented as $(X,Y)$ in exactly one way. Meanwhile, every hand with an exactly $10$-long s-f can be represented as $(X,Y)$ in exactly two ways, and every hand with an exactly $11$-long s-f can be represented as $(X,Y)$ in exactly three ways, etc. Therefore:
$$5 \times {52 - 9 \choose 13 - 9} = N_9 + 2 N_{10} + 3 N_{11} + 4 N_{12} + 5 N_{13}$$
where $N_j$ is the no. of hands with an exactly $j$-long s-f.
By a completely analogous argument but this time considering $(W,Z)$ where $|W| = 10$ and the contents form a $10$-long s-f, and $|Z| = 3, W \cap Z = \emptyset$, we have
$$ 4 \times {52 - 10 \choose 13 - 10} = N_{10} + 2 N_{11} + 3 N_{12} + 4 N_{13}$$
SubtractingThe difference between the two, we have gives the desired count, i.e. no. of hands with $9$-long (or longer) s-f in spades:
$$N_9 + N_{10} + N_{11} + N_{12} + N_{13} = 5 \times {52 - 9 \choose 13 - 9} - 4 \times {52 - 10 \choose 13 - 10} = 571130 \,\,\,\,\,\,\square$$