In short, the probability of a 7-card straight when drawing 7 random cards from a standard deck of 52 is $0.000979$.
To calculate this value, we note that all 7-card hands are equally likely, of which there are ${52 \choose 7} = 133,784,560$ possibilities.
Next, we compute the number of 7-card straights. Ignoring suit, we note that there are $8$ possible straights (starting with {A, 2, 3, 4, 5, 6, 7} through {8, 9, 10, J, Q, K, A}). For each card in the straight, there are 4 possibilities for the suit, such that there are $4^7 = 16384$ ways to assign the suits to the 7 cards. However, $4$ of these suit assignments yield straight flushes (all clubs, all diamonds, etc.), so the actual number of suit assignments that can yield a straight (but not a straight flush) is $16384 - 4 = 16380$.
Putting all this together, there are $8 \times 16380 = 131,040$ possible 7-card straights out of $133,784,560$ possible 7-card hands, yielding a probability of $\approx 0.000979$.
As @Glen_b suggested, it would be a good idea to summarize the comments part in an answer. I'll do that and also give an alternative for the formula related to the probabilistic point of view at the end of the answer. The apparent contradiction between the two computations came from this line :
the probability of having the Ace of Spades is:
$$\frac{1}{52}+\frac{1}{51}+\frac{1}{50}+\frac{1}{49}+\frac{1}{48}=\frac{433507}{4331600}\approx\frac{1}{10}>\frac{5}{52}$$
The idea behind this computation was good but the logic was flawed. If we sum the probability of the Ace of Spade drawn exactly at the k th draw with k going from 1 to 5, we have the probability we want. But $\frac{1}{51}$, for example, does not represent the probability that the ace of spade is exactly drawn at the second attempt but the probability that the ace of space is drawn at the second attempt given that it has not been drawn at the first one.
AJS finally found the right formula
I got it! $1/52+((51/52)∗(1/51))+⋯+((51/52)∗(50/51)∗(49/50)∗(48/49)∗(1/48))=5/52$
With the idea that the probability to exactly draw the ace of spade at the $k$ th trial is the probability to not draw the ace of spade during previous attempts.... $$(51/52)*(50/51)...(52-k+1)/(52-k+2)$$...multiplied by the probability to draw the card at the $k$ th attempt $$1/(52-k+1)$$
A general rule of thumb that I have to avoid this kind of error is to be cautious when it comes to adding probabilities. If you can turn your problem around to avoid additions to the profit of multiplications, this is less prone to error.
A more classic approach giving a shorter path would have been to consider that the probability of having the Ace of Spades after 5 draw is 1-(the probability to not draw it with 5 draws) which gives, as you know : $$1-(51/52)*(50/51)*(49/50)*(48/49)*(47/48)=1-47/52=5/52$$
Best Answer
This program simulate you game
The error is small so your formula it's probably right