$n=10$:

Attempt 1.1: Expansion of arctan(1) [3 terms]

Yeah, this might not be a great idea. We know that

$$4\arctan1=\pi=4\sum_{n\ge0}\dfrac{(-1)^n}{2n+1}$$

So we can approximate this with three terms to get

$$4\arctan1\approx4\left(1-\dfrac13+\dfrac15\right)\\=4\times\dfrac{13}{15}=4\times\dfrac{7+6}{3\times5}$$which is not accurate at all. ($=3.4\overline6$)

Attempt 1.2: Expansion of arctan(1) [4 terms]

We can now try this with 4 terms with

$$4\arctan1\approx4\left(1-\dfrac13+\dfrac15-\dfrac17\right)\\=4\times\dfrac{76}{105}=4\times\dfrac{6(10+1+2)}{3\times5\times7}$$which is also not accurate.

Attempt 1.3: Expansion of arctan(1) [5 terms]

Using 5 terms, we can try to get to 315 in our denominator and 263 in our numerator. The only problem is I am unsure how to. Here is my attempt (since we need the 4 to multiply our expansion):

$$263:\text{ }2^8+7\text{ (numbers left: }1,3,5,6,9,10)\\315:\text{ (unable to do, closest was }316)\text{ }3^5+7\times9+10=316$$and now we do$$4\times\dfrac{263}{316}\approx3.3291$$which is actually more accurate than if I had managed to get 315^[1], and we can get this more accurate with using up 6 and 1, which I managed to not use to get$$4\times\dfrac{263}{323}=4\times\dfrac{2^8+7}{3^5+7\times9+10+1+6}\approx3.25696594$$which is even more accurate than before (and in my testing with expansions, is the most accurate I managed to get it) but still not close enough to even be considered probably. I’ll come back to this in the future.

Attempt 2: Unoriginality (original method by 2012rcampion)

Yeah, we’re using $\dfrac{355}{113}$ again. However, this time, we can use all of the numbers:

$$\dfrac{355}{113}=\left(3+\dfrac{2\times8}{6\times4\times5-7}\right)(10-9)\approx3.141592920\dots$$which yet again has an error of $2.67E-7$.

However, I will try to provide a fully original solution for $n=10$ in the future.

^[1]As I finish up writing this post, I just realized I’m stupid. I could have just subtracted 1 from 10 to get $7\times9+10-1=7\times9+9=9(7+1)=8\times9$ which would have been all that I would have needed to do to get the $315$ I wanted. But hey, I’ve already found a better solution for my expansion of $4\arctan1$ already, so it doesn’t matter.

Edit 1: I have a solution for $n=17$:

$n=17$:

Attempt 1: Small digit approximation method.

We get this from using the small digit approximation for $\pi$: [source]

$$.1^{-2/3}+(4/.5)^{-6}-.7-.8$$

So what do we do?

First off, we can rewrite this as$$(1/10)^{-2/3}+8^{-6}-(7/10)-(8/10)$$Now, rewrite $-2/3$ as $(4-6)/3$ and $7/10$ as $7/(12-2)$ so now it is$$\left(\dfrac1{10}\right)^{(4-6)/3}+8^{-6}-\dfrac7{12-2}-\dfrac8{10}$$Now, to get our -6, we replace that with 5-11 (which is the only replacement that I found that I was able to make that would not require me to restart this) and finally, we use $\dfrac{17-13}{14-9}=\dfrac45$ to rewrite our approximation for $n=17$ as (without using 16 or 15)$$\left(\dfrac1{10}\right)^{(4-6)/3}+8^{5-11}-\dfrac7{12-2}-\dfrac{17-13}{14-9}$$which now the approximation is improved to just be $5.28E-9$ less than $\pi$, my most accurate solution so far.