Sizes in the Dungeons & Dragons scale with the 5' side area. Taking a size factor F, this can be put into a formula: \$F*5'=L_{side}\$ and \$A=L_{side}^2\$. This makes a square occupy 25 square feet.
Factor 1
Medium and small creatures are a factor 1, they have a side length of 5' and occupy a full square.
Factor 0.5
A tiny creature is half of that for 2.5' side length. This means that 4 fit into a 5'square. Typical Examples are cats for the mid range, but we should check one factor lower to see how small we can get.
Factor 0.25
If you get smaller, you leave where D&D5 has terms for. In earlier edition you would have got a "Diminuitive" creature, occupying a square with 1.25' side length. 16 fit into the 5' square, occupying 1.5625 square feet. This is by far not RAW, but can help to determine the size.
Interpretation
We know that beings are always a bit smaller than the square they occupy - the main difference between small and medium being height. But how much space do you need to effectively control an area of 5' by 5'?
First of all, you need to effectively reach from the center of your occupied area to the one next to you, because you need that to attack a creature in the neighboring square. So, in best case it is once the creatures \$L_{side}\$, in the worst case this demands us to have \$L_{side}\times\sqrt{2} = L_{Arm}\$.
Virtuvian Man
Now, time to pull out the virtuvian man for a medium creature: \$2\times L_{Arm}+L_{Sholderwidth}=L_{Height}\$ and \$4 \times L_{Shoulderwidth}=\frac34\times(2\times L_{Arm})=L_{Height}\$
Shifting the numbers around this gives us a size of \$L_{Side}\$ for the average size and an absolute maximum height of \$L_{Height}=\frac32\sqrt2 L_{Side}=2.12\times L_{Side}\$
So, for the average we get 5' and as absolute upper limit of a medium creature, we gain 10.24 feet.
Virtuvian Spider?
Now, assuming this 'Virtuvian Man' can be applied equally to smaller creatures, especially the Tiny Category, we get \$L_{Side}\$ as the average height of such a being, \$\sqrt2 L_{Side}\$ for maximum. Well, for a spider, we speak about the absolute length with stretched out legs. Assuming it is more or less just a scaled up Black Widdow, then each of the 3 parts of a leg is as long as the torso, for a total of 5 to 7 roughly equall length pieces (2*Leg+Torso=5*Torso) forming the length.
Conclusion
So, the spider is like \$2.5'\text{ to }2.5\sqrt2=3.5'\$ in diameter, for a torso size of \$6''\text{ to }8.5''\$ with legs of at least 1 foot long. Which is a scary big spider, considering the largest spider we know today Theraphosa blondi has a diameter of 1 foot and would perfectly well fit into the size limits of a tiny creature.
Since there are no smaller sizes, this is the very upper limit of our spider's size - it could very well be much smaller!
With previous generations in mind:
If taking diminuitive size into account, the smallest tiny spider would have a diameter of 15" to 21.2", being larger than the Bird Eating Spider.
Diminuitive spiders would range between 7" and those 21", filling in exactly the gap of that Spider Beast,
Anything smaller than 7" would have ben a "fine spider". Not fine as good but as D&D 3.5 size category.
But again, as D&D5e stopped with that kind of below tiny categorization (like fitting 64 tiny spiders into a 25 square feet area before forcing you to apply the swarm template for more creeping horror): Your \$Tiny\text{ }Spider^{TM}\$ only has an upper limit of about 3.5 feet fully stretched with no real limit to the underside.