That $G(\psi,\zeta)^2 = \psi(-1)p$ is pure algebra, so it holds in $\mathbf C$ or $\mathbf C_p$ or any other field not of characteristic $2$ that contains a nontrivial $p$th root of unity.

You could write down a $p$-adic formula for your quadratic Gauss sum using the Gross–Koblitz formula.

First let's normalize the link between your nontrivial $p$th root of unity and your choice of $\pi$ such that $\pi^{p-1} = -p$. To each $\pi$ there is a unique nontrivial $p$th root of unity $\zeta$ such that $\zeta \equiv 1 + \pi \bmod \pi^2$, where the congruence means $\lvert\zeta - (1 + \pi)\rvert_p \leq \lvert\pi\rvert_p^2$, or equivalently $\lvert\zeta - (1 + \pi)\rvert_p < \lvert\pi\rvert_p$ since $\mathbf Q_p(\pi) = \mathbf Q_p(\zeta)$. Write the $\zeta$ fitting that congruence mod $\pi^2$ as $\zeta_{\pi}$.

Every character of $\mathbf F_q^\times$ with values in $\mathbf C_p$ is a power of the Teichmüller character $\omega_q$ (interpret $\mathbf F_q$ as $\mathbf Z_p[\zeta_{q-1}]/(p)$). For the Gross–Koblitz formula it is convenient to write characters of $\mathbf F_q^\times$ as powers of $\omega_q^{-1}$, say as $\omega_q^{-k}$ for $0 \leq k < q-1$. The quadratic character $\psi$ of $\mathbf F_q^\times$ is $\omega_q^{(q-1)/2} = \omega_q^{-(q-1)/2}$, so $k = (q-1)/2$.
Let the base $p$ expansion of $k$ be $d_0 + d_1p + \cdots + d_{f-1}p^{f-1}$. When $k = (q-1)/2 = (p^f-1)/2$, all of its base $p$ digits are $(p-1)/2$, so the sum of the base $p$ digits is $f(p-1)/2$.

The Gross–Koblitz formula for the quadratic character $\psi$ says
$$
-G(\psi,\zeta_\pi) = \pi^{f(p-1)/2}\Gamma_p\left(\frac{(p-1)/2}{q-1}\right)^f,
$$
where $\Gamma_p$ is Morita's $p$-adic Gamma-function. Note the minus sign on the left side: normalizing Gauss sums with an overall minus sign is reasonable for various purposes, like here and in the Hasse–Davenport relation. On the right side of the formula above,
$\pi^{f(p-1)/2}$ is a square root of $\pi^{f(p-1)} = (-p)^f = (-1)^fq$.

$\newcommand\sgn{\genfrac(){}{}}$In the special case $q = p$ (so $f = 1$), you're working with the classical quadratic Gauss sum for $\mathbf F_p$ and the Legendre symbol. In this case
$$
-G\left(\sgn\cdot p,\zeta_\pi\right) = \pi^{(p-1)/2}\Gamma_p\left(\frac{1}{2}\right),
$$
where $\pi^{(p-1)/2}$ is a square root of $\pi^{p-1} = -p$. For $p > 2$ it is known that $\Gamma_p(1/2)^2 = -\sgn{-1}p$, so if you square the right side above then you get $\pi^{p-1}\Gamma_p(1/2)^2 = -p(-\sgn{-1}p) = \sgn{-1}p p$, which is the formula for the square of the mod $p$ quadratic Gauss sum that I mentioned at the start of this answer (when $q = p$).