Assuming by the hypergeometric function, you mean the trace of Frobenius the hypergeometric sheaves, there is a nice relation.

First I will mention the right way to move a character to a higher degree finite field. If $A$ is a character of $\mathbb F_q^\times$, then the corresponding character of $\mathbb F_{q^n}^\times$ is $A$ composed with the norm from $\mathbb F_{q^n}$ to $\mathbb F_q$. This is correct because both characters correspond to the same sheaf.

Thus a hypergeometric function $2F1(A,B,C|x)$ over $\mathbb F_{q^2}$ corresponds to a hypergeometric function $2F1(A',B',C'|x)$ over $\mathbb F_{q^2}$ because they come from the same sheaf, if $A$ corresponds to $A'$, $B$ corresponds to $B'$, and $C$ corresponds to $C'$.

But what does this say about the hypergeometric function? Well the function is just the trace of Frobenius. Since $Frob_{q^n}=\left(Frob_q\right) ^n$, we can use an identity about the traces of powers of matrices. Since the sheaf has rank $2$, the matrices are two-dimensional, and we have the identities:

\[ \operatorname{tr}(M^2) = \operatorname{tr}(M)^2 - 2 \det(M)\]

\[ \operatorname{tr}(M^3) = \operatorname{tr}(M)^3 - 3 \det(M) \operatorname{tr}(M)\]

and so on, using the Chebychev polynomials. Here $ \operatorname{tr}(M)$ corresponds to the hypergeometric function of a point over $\mathbb F_q$, $ \operatorname{tr}(M^2)$ to the corresponding function of the same point over $\mathbb F_{q^2}$, and $ \operatorname{tr}(M^3)$ to the corresponding function of the same point over $\mathbb F_{q^3}$. It remains to give a nice formula for the determinant.

I'm pretty sure there is one, although I don't know exactly what it is. It should be something like $p^3 A\left(\frac{x}{x-1}\right)B\left(\frac{x}{x-1}\right)C(x-1)$. I'm not sure precisely what notation you are using. Without an exact formula, you still get a relation among $\mathbb F_p$, $\mathbb F_{p^2}$, and $\mathbb F_{p^3}$, or between any three finite fields, using the symmetric functions.