This question has a somewhat trivial answer because the polynomial $x^{2^s}$ requires $s$ multiplications, so you can just take $h = x_1^{2^{f(n)}}$. This is one of the reasons why in algebraic complexity we almost always talk about families with polynomially bounded degree.
For a less trivial statement, we can use the same strategy as in the Boolean settings, but instead of simple counting, in algebraic complexity we use dimension counting.
Main idea: in a circuit computing a non-constant function has $s$ gates, then it has at most $s$ constants. By fixing the structure of the circuit and varying the constants we see that the set of all polynomials computed by the fixed circuit (which is a Zariski constructible set by Chevalley's theorem) has dimension at most $s$. There is only finite number of circuits of complexity $s$, so the set of all polynomials with complexity $s$ has dimension at most $s$. Comparing $s$ with the dimension of the set of polynomials with given degree and number of variables, we can get a statement about the existence of hard polynomials.