If you think about how you might go about writing efficient code to enumerate all the straight flushes, you can develop a mathematical formula--which is the ultimate efficiency! I will not in fact bring you all the way to such a formula, because it's a nuisance. Instead, I will show how to short-circuit both the complicated combinatorial maneuvers and computationally-intensive brute-force searches to find a happy medium in which you can be reasonably certain of getting the correct answer while not having to work too hard to do so.
First, note that it's not possible for a seven-card hand to hold straight flushes (SF) of two different suits, because (apart from a possible Joker) there have to be at least four suited cards in the SF. Therefore we need only count the straight flushes of a given suit, and multiply that count by four to get the total.
Pick a suit, once and for all. Let's divide the problem of finding all the SFs in that suit according to how many cards of that suit (counting the Joker) actually appear in the hand. Since a SF consists of five cards, the (mutually exclusive) possibilities are $7$, $6$, and $5$ cards. Let's count each case. To simplify the notation, let the ranks be 1 (the ace also written as A), 2, ..., 9, T=10, J=11, Q=12, and K=13. The ace can count as rank 14 if that helps to form a SF.
With $5$ cards, they are the SF. Without the Joker, their sorted ranks can be 12345, 23456, ... through TJQKA, of which there are $10$ possibilities. If one of them is the Joker, we have to take care not to double-count. Sort them lexicographically, replacing the Joker (*) with the smallest value that can create a SF. For instance, 3456* is interpreted as 23456 rather than 34567. In this fashion we obtain five more possible ways of forming 12345 (namely, *2345, 1*345, 12*45, 123*5, and 1234*), but only four additional ways of forming the other SFs. (E.g., the five SFs beginning with 3 are 34567, *4567, 3*567, 34*67, and 345*7: the set 3456* would be interpreted as *3456, which begins with 2!) That produces $10 + 4(10) + 1 = 51$ distinct possibilities. There are, independently of these possibilities, $\binom{39}{2}$ ways to select the cards in one of the other three suits.
With $6$ cards, the counting gets more complicated but is not fundamentally different. Note that we cannot blithely count the SFs without a Joker and then multiply that by $6$ to account for the six possible positions a Joker could substitute for. (This is where other attempts at a solution have foundered.) The problem is that this counts some hands multiple times. As an example, consider the hands 89TJQK and 8TJQKA. By substituting a Joker (*) for the 9 in the first and for the Ace in the second, we obtain identical hands 8TJQK*. Instead, we can order the SFs as before, by the start of the lowest SF that can be formed from the cards, but this time we have to multiply the counts by the number of value which that spare sixth card can have without creating a SF with a lower starting value. The resulting count is $356$. It breaks down into $49$ ways to product a SF beginning with an ace, $35$ SF's beginning with 2, and $34$ each of all the rest. As before, this value of $356$ has to be multiplied by the $\binom{39}{1}$ ways to select the seventh card from the other $39$ cards not in the suit.
With $7$ cards, the counting is further complicated because there are two spare cards floating around, but no new ideas are needed. The actual count is $1066$. It breaks down, ordered by the start of the SF, into $176, 105, 99,$ and $98$ each of the remaining SFs (beginning with 4 or greater).
What's the point of just quoting these counts? Well, if we wish, we can obtain the two difficult counts (for $6$ and $7$ suited cards) by brute force enumeration. The number of seven-card hands that can be drawn from a Joker and only the $13$ cards of one suit is a mere $\binom{13+1}{7} = 3432$. Instead of having to search through hundreds of millions of possibilities ($\binom{53}{7} = 154\ 143\ 080$), we can perform an almost instantaneous search, once. The searches over six-card and five-card hands are even shorter.
The total number of SFs, allowing one Joker, therefore is
$$k = 4\left(51\binom{39}{2} + 356\binom{39}{1} + 1066\right) = 210\ 964.$$
From this we obtain the chance of drawing a straight flush as
$$\frac{k}{\binom{53}{7}} = \frac{210\ 964}{154\ 143\ 080} = \frac{4057}{2\ 964\ 290} = 0.00136883\ldots\ .$$
Because it's rather a nuisance to work out the values $356$ and $1066$ with formulas, this solution offers a nice combination of tools: use a little combinatorial reasoning to (greatly) reduce the computation, and then use the computer to avoid working too hard.
Readers who have come this far will be pleased to note that this answer agrees with the one posted earlier by Kodiologist.
