Here is a negative answer to your first question:
$\overline{\mathbb{Z}}/p\overline{\mathbb{Z}}$ is not a field and not even an integral domain, because $p\overline{\mathbb{Z}}$ isn't a prime ideal.
Consider the factorisation $p=\sqrt{p}^2$, of course $\sqrt{p} \in \overline{\mathbb{Z}}$
Now we will assume that $p\overline{\mathbb{Z}}$ is a prime ideal of $\overline{\mathbb{Z}}$.
Since $\sqrt{p}^2=p \in p\overline{\mathbb{Z}}$, we must have $\sqrt{p}\in p\overline{\mathbb{Z}}$. This means that $\sqrt{p}=ap$ for some $a\in \overline{\mathbb{Z}}$, which would imply $p=\sqrt{p}^2=a^2p^2 \implies 1=a^2p$, which is a contradiction because $p$ is not a unit of $\overline{\mathbb{Z}}$.
Addition: A positive answer to your second question.
Let's assume $\overline{\mathbb{Z}}/p\overline{\mathbb{Z}}$ is finite and let $m<\infty$ be its order. It is obvious that $m>1$. We will show that $p^n$ divides $m$ for arbitrarily high $n$.
Pick an arbitrary number field $K$ of degree $n=[K:\mathbb{Q}]$ and consider the obvious map $\mathcal{O}_K \to \overline{\mathbb{Z}}/p\overline{\mathbb{Z}}, x \to x +p\overline{\mathbb{Z}}$.
Let $x \in \mathcal{O}_K$ be in the kernel of this map, so $x=ap$ for some $a \in \overline{\mathbb{Z}}$. We have $K \ni x/p=a \in \overline{\mathbb{Z}}$, so $a \in \overline{\mathbb{Z}} \cap K = \mathcal{O}_K$, which means the kernel of this map is $p\mathcal{O}_K$.
This implies that $\mathcal{O}_K/p\mathcal{O}_K$ injects into $\overline{\mathbb{Z}}/p\overline{\mathbb{Z}}$ and $p^n=\#\mathcal{O}_K/p\mathcal{O}_K \mid \#\overline{\mathbb{Z}}/p\overline{\mathbb{Z}}=m$.
Picking an arbitrary sequence of number fields $K_1,K_2,...$ with $[K_i:\mathbb{Q}]\to \infty$ (which certainly does exist) leads to a contradiction.
Best Answer
Let me expand Ethan Alwaise's comment into an answer.
Let $L$ be a number field and $\alpha\in \mathcal O_L$ be a square modulo every prime. Suppose by contradiction that $\alpha$ is not a square. Then the polynomial $x^2-\alpha$ is irreducible in $L[x]$. Hence, the field extension $L(\sqrt{\alpha})/L$ has degree 2. This implies that the Galois group of such extension is $C_2$. Hence, by Chebotarev there exist infinitely many primes of $L$ that are inert in $L(\sqrt{\alpha})$. Choose one of these primes $p$ large enough. Then by Dedekind-Kummer the splitting of $p$ in $L(\sqrt{\alpha})$ is governed by the factorization of $x^2-\alpha$ modulo $p$. But this is a product of two linear factors by hypothesis, leading to a contradiction.