In the context of Carlitz polynomials, the analogue is the classical quadratic reciprocity law in ${\mathbf F}_q[T]$ (for odd $q$, of course). For a monic irreducible $\pi$ in ${\mathbf F}_q[T]$, and any $A$ in ${\mathbf F}_q[T]$, let $(\frac{A}{\pi})$ be $1$ if $A \equiv \Box \bmod \pi$ and $A \not\equiv 0 \bmod \pi$, $-1$ if $A \not\equiv \Box \bmod \pi$, and $0$ if $A \equiv 0 \bmod \pi$. Then the main law of quadratic reciprocity in ${\mathbf F}_q[T]$ is that for distinct monic irreducible $\pi$ and $\widetilde{\pi}$,
$$
\left(\frac{\widetilde{\pi}}{\pi}\right) = (-1)^{({\rm N}\pi-1)/2 \cdot ({\rm N}\widetilde{\pi}-1)/2}\left(\frac{\pi}{\widetilde{\pi}}\right),
$$
where ${\rm N}(f) := |{\mathbf F}_q[T]/(f)| = q^{\deg f}$. This is due to Dedekind, and can be found in Mike Rosen's Number Theory in Function Fields with a proof that is much simpler than quadratic reciprocity in $\mathbf Z$ and doesn't mention Carlitz polynomials at all. But it can be explained using Carlitz extensions of ${\mathbf F}_q(T)$ in exactly the same way quadratic reciprocity in $\mathbf Z$ can be explained using cyclotomic extensions of ${\mathbf Q}$.
Let's recall first what is done in the case of the integers. For an odd prime $p$ the cyclotomic extension ${\mathbf Q}(\zeta_p)$ has a cyclic Galois group over $\mathbf Q$ of order $p-1$ and thus it contains exactly one quadratic extension of $\mathbf Q$ (corresponding to the subgroup of squares in the Galois group). This quadratic extension is ${\mathbf Q}(\sqrt{p^*})$, where $p^* = (-1)^{(p-1)/2}p$. For any odd prime $q$ other than $p$, we will compute the Frobenius element $\text{Frob}_q({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$ in two ways. We will interpret $\text{Gal}({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$ as $\{\pm 1\}$, which can be done in a unique way (all groups of order 2 are uniquely isomorphic to each other), so the Frobenius element is 1 or $-1$.
(1) By knowledge of how primes split in quadratic extensions of ${\mathbf Q}$, $q$ splits in ${\mathbf Q}(\sqrt{p^*})$ iff $X^2 - p^*$ splits mod $q$, so this Frobenius element is $(\frac{p^*}{q})$.
(2) From the standard isomorphism of $\text{Gal}({\mathbf Q}(\zeta_p)/{\mathbf Q})$ with $({\mathbf Z}/(p))^\times$, $\text{Frob}_q({\mathbf Q}(\zeta_p)/{\mathbf Q})$ is $q \bmod p$. The restriction of $\text{Frob}_q({\mathbf Q}(\zeta_p)/{\mathbf Q})$ to ${\mathbf Q}(\sqrt{p^*})$ is $\text{Frob}_q({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$, and the image of $q \bmod p$ under the natural restriction map $\text{Gal}({\mathbf Q}(\zeta_p)/{\mathbf Q}) \rightarrow \text{Gal}({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$ is $(\frac{q}{p})$. Therefore $\text{Frob}_q({\mathbf Q}(\sqrt{p^*})/{\mathbf Q}) = (\frac{q}{p})$.
By (1) and (2), $(\frac{p^*}{q}) = (\frac{q}{p})$, and this is the main law of quadratic reciprocity after we recall that $p^* = (-1)^{(p-1)/2}p$ and use the supplementary law $(\frac{-1}{q}) = (-1)^{(q-1)/2}$.
Now let's turn to the Carlitz setting. Pick a monic irreducible $\pi$ in ${\mathbf F}_q[T]$, where $q$ is odd. Let $\Lambda_\pi$ be the set of roots of the Carlitz polynomial $[\pi](X)$. Set $K = {\mathbf F}_q(T)$, so $K(\Lambda_\pi)/K$ is a Galois extension with Galois group isomorphic to $({\mathbf F}_q[T]/\pi)^\times$ by using the action of Carlitz polynomials on $\Lambda_\pi$. Any monic irreducible $\widetilde \pi$ other than $\pi$ is unramified in $K(\Lambda_\pi)$, and $\text{Frob}_{\widetilde{\pi}}(K(\Lambda_\pi)/K) = \widetilde{\pi} \bmod \pi$.`
Letting $d = \deg \pi$, the group $({\mathbf F}_q[T]/\pi)^\times$ is cyclic of odd order $q^d-1$, so it has a unique subgroup of index 2 (the subgroup of squares). Therefore $K(\Lambda_\pi)$ has a unique quadratic subextension of $K$ inside it. It turns out to be $K(\sqrt{\pi^*})$, where $\pi^* = (-1)^{({\rm N}\pi-1)/2}\pi$.` (EDIT: This is explained below.)
If you run through the above proof of quadratic reciprocity in $\mathbf Z$ using the analogous constructions in the Carlitz setting that I describe above, then you'll obtain $$\left(\frac{\pi^*}{\widetilde{\pi}}\right) = \left(\frac{\widetilde{\pi}}{\pi}\right)$$ in exactly the same way that one obtains $(\frac{p^*}{q}) = (\frac{q}{p})$, and the main law of quadratic reciprocity in ${\mathbf F}_q[T]$ is an unraveling of this equation once you recall the definition of $\pi^*$ and use the supplementary law $(\frac{-1}{\widetilde{\pi}}) = (-1)^{({\rm N}\widetilde{\pi}-1)/2}$.
EDIT: I wrote in passing above that $K(\Lambda_\pi)$ contains a square root of $\pi^* := (-1)^{({\rm N}\pi-1)/2}\pi$. This is analogous to ${\mathbf Q}(\zeta_p)$ containing a square root of $p^* = (-1)^{(p-1)/2}p$, but explaining the containment is one place where simple analogies between ${\mathbf Q}$ and ${\mathbf F}_q(T)$ can break down. Classically there are several ways of showing $\sqrt{p^*}$ lies in ${\mathbf Q}(\zeta_p)$.
(1) Ramification. The unique quadratic field in ${\mathbf Q}(\zeta_p)$ ramifies only at $p$, and there turns out to be just one quadratic extension of ${\mathbf Q}$ ramified only at $p$.
(2) Gauss sums. Define $G = \sum_{a \bmod p} (\frac{a}{p})\zeta_p^a$, which by construction lies in ${\mathbf Q}(\zeta_p)$. Show $G^2 = p^*$.
(3) Calculating a norm in a second way. Setting $X = 1$ in the identity $X^{p-1}+\cdots+X+1 = \prod_{i=1}^{p-1}(X-\zeta_p^i)$ gives us $p = \prod_{i=1}^{p-1} (1 - \zeta_p^i)$. This says $p = {\rm N}_{{\mathbf Q}(\zeta_p)/{\mathbf Q}}(1- \zeta_p)$. The product of the terms at $i$ and $p-i$ in the product is $-1$ up to a square factor (because $\zeta_p$ is a square of some $p$th root of unity). Therefore $p$ is $(-1)^{(p-1)/2}$ times a square in ${\mathbf Q}(\zeta_p)$, so $(-1)^{(p-1)/2}p$ is a square in the $p$th cyclotomic field.
If we try to adapt these methods to find the unique quadratic extension of $K = {\mathbf F}_q(T)$ inside $K(\Lambda_\pi)$, the first two do not work directly.
(1) There is not just one quadratic extension of $K$ ramified only at $\pi$ among the "finite" places (those places other than the place associated to $1/T$). One choice is $K(\sqrt{\pi})$ and another is $K(\sqrt{c\pi})$ where $c$ is a nonsquare in ${\mathbf F}_q^\times$.
(2) If we define $G = \sum_{A \bmod \pi} (\frac{A}{\pi})[A](\lambda)$ for any fixed choice of nonzero $\lambda$ in $\Lambda_\pi$, then $G = 0$ if ${\rm N}(\pi) > 3$ (not if ${\rm N}(\pi) = 3$). To prove $G=0$ we exploit additivity of Carlitz polynomials: the coefficients $(\frac{A}{\pi})$ are $\pm 1$, so
$$
\sum_{A \bmod \pi} \left(\frac{A}{\pi}\right)[A](\lambda) =
\left[\sum_{A \bmod \pi}\left(\frac{A}{\pi}\right)A\right](\lambda),
$$
and the polynomial inside the brackets on the right only matters mod $\pi$ since its Carlitz action is being applied to $\lambda$, a root of $[\pi](X)$. Therefore the vanishing of $G$ is the same as the vanishing of $\sum_{A \bmod \pi} (\frac{A}{\pi})A \bmod \pi$, and Darij Grinberg gives a proof of that in his comments to this answer. (When ${\rm N}(\pi) = 3$ the sum $G$ is $2\lambda \not= 0$.)
The idea of the third method does carry over to the Carlitz setting.
(3) Starting from the factorization $[\pi](X)/X = \prod_{A \not\equiv 0 \bmod \pi} (X - [A](\lambda))$, set $X = 0$ (not $X = 1$: the roots of $[\pi](X)$ are more analogous to $1 - \zeta_p^i$ instead of to the $p$th roots of unity themselves) and get $\pi = \prod_{A \not\equiv 0 \bmod \pi} [A](\lambda)$. This says $\pi = {\rm N}_{K(\Lambda_\pi)/K}(\lambda)$.` The product of the terms at $A$ and $-A$ is $[A](\lambda)[-A](\lambda) = -[A](\lambda)^2$ because $[-A](X) = [-1]([A](X)) = -[A](X)$. Therefore up to a square factor in $K(\Lambda_\pi)$, $\pi$ equals $(-1)^{({\rm N}\pi- 1)/2}$, so $(-1)^{({\rm N}\pi- 1)/2}\pi$ is a square in $K(\Lambda_\pi)$.