In a very flattering note @saulspatz has asked me to weigh in on this issue. But many of the commenters to your question have already evinced rather greater familiarity with deep $p$-adic matters than I am in possession of. Nonetheless, here’s my feather’s worth.

My first reaction to your question is, “There can’t be such a thing as $p$-adic $\pi$.” At the outset, however, let me warn you that any time an older scientist says that something is impossible, it’s time for young scientists to prove the old geezer (or old hag) wrong. Just think of Kelvin’s preposterous (in light of the geologic knowledge in his time) estimate of the earth’s (and worse, of the sun’s) age.

What archimedean phenomenon would one want to form an analogy to, for defining a $p$-adic $\pi$? Certainly not the ratio, somehow, of the circumference (or area) of a circle to the radius. For there are no circles in $\Bbb Q_p$, except disks, and they don’t have circumferences.
You might try comparing the “interior” $\{z\in\Bbb Q_p:|z|<1\}$ of the unit disk $D_1=\{z\in\Bbb Q_p:|z|\le1\}$, but that’s not its interior, and what would naively be thought to be the “circumference”, the set-difference of the two, has $p-1$ times the measure of the “interior”. No good for anything like the real $\pi$, especially when you see that the same construct delivers different ratios under finite extensions of the field $\Bbb Q_p$.

Anyone would run screaming from that approach, and hope that the first nonzero root of the sine function, or the first nonzero argument $z$ of the complex function $z\mapsto e^{iz}$ to give the value $1$, might help.
Even ignoring the question that there can be no good $p$-adically defined notion of firstness, neither of these functions is defined (or, apparently, definable) far enough from zero to be anything other than one-to-one. In the case of the sine function, the series expansion that you know makes perfectly good sense in a neighborhood of $0$, but on its $p$-adic domain of convergence, it’s one-to-one, so has just the one root $0$.

You might think that the exponential offered more hope, but its domain of convergence is the same, $\{z:|z|<p^{-\frac1{p-1}}\}$. (At this point, the multiplicative absolute value confuses me, and I have to use the additive valuation $v_p:\Bbb Q_p\to\Bbb Z\cup\{\infty\}$. It satisfies $v(zz')=v(z)+v(z')$, $v(z+z')\ge\min(v(z),v(z'))$, and $v(0)=\infty$, $v(p)=1$. ) In my preferred notation, the condition for convergence of the exponential is $v(z)>\frac1{p-1}$. But no matter how you look at it, the exponential is one-to-one on that domain, has no other zeros.

Oh well, you say, how about the logarithm? We know that in the right domain of definition, we have $\log(i)=i\pi/2$. Why can’t we use that? But the glory of complex analysis is that the logarithm is *not a homomorphism*. It tries to satisfy the rule $\log(zz')=\log z+\log z'$ all right, but the logarithm is not defined on a group.

I suppose you can say that the glory of $p$-adic analysis is that the logarithm *is* a homomorphism, and it’s defined on the huge group $\{z\in\Bbb C_p: v(z-1)>0\}$, the principal units of $\Bbb C_p$. The codomain is $\Bbb C_p$, and it’s an easy exercise to show that the log is surjective (!). But for every $z\in\Bbb C_p$, the inverse image $\log^{-1}(z)$ is infinite, because the kernel of the *homomorphism* $\log$ is the set of $p$-power roots of unity, an infinite subgroup of the domain group.

You offer another suggestion: maybe some other archimedean series can be used for defining a $p$-adic $\pi$? I don’t think that the $\arctan$ series will help, $\arctan(x)=x-x^3/3+x^5/5-\cdots$. In the real world you can set $x=1$ there, but the series is $p$-adically divergent. What would $\arctan(p)$ a anyway? But one might look at that…

Philosophically speaking, I think you’re chasing a will-o’-the-wisp. I think that $\pi$ is intrinsically an archimedean object, and to ask for a $p$-adic analog is to ask for a contradiction in terms.