Here is a related statement which can be proven "algebraically" (as opposed to going to the boolean world of #P/poly).
Suppose $f(x_1,\ldots,x_n,y_1,\ldots,y_m) \in \text{VNP}$ for $m = \text{poly}(n)$. Then for any $e_1,\ldots,e_m$, the "coefficient" of $y_1^{e_1}y_2^{e_2} \cdots y_m^{e_m}$ in $f$ (a polynomial in $x_1,\ldots,x_n$) is also in $\text{VNP}$.
Proof sketch: We will prove the statement for $m=1$, the main statement would then follow.
Suppose the degree of $y$ in $f$ is strictly less than $D = 2^d$, for some integer $d$ (pick the smallest one).
Note that for any polynomial $g(y)$ of degree $< D$, and any $0 \leq e_1 < D$, we have that for every choice of distinct constants $a_0, a_1, \ldots, a_{D-1}$, there exists constants $\beta_0,\ldots,\beta_{D-1}$ (independent of $g$) such that the coefficient of $y^e$ in $g(y)$ is given by $\sum_{0 \leq i \leq D-1} \beta_i g(a_i)$. This is just univariate interpolation.
Next, note that we can design a polynomial $c_e(a)$ of degree at most $D$, such that $c_e(a_i) = \beta_i$ for each $i$.
Now, here is a "$\text{VNP}$-expression" for the coefficient of $y^e$ in $f(x_1,\ldots,x_n,y)$.
$$ \sum_{i_1,\ldots,i_d \in \{0,1\}} c_e(val(i_1,\ldots,i_d)) f(x_1,\ldots,x_n,val(i_1,\ldots,i_d)),$$
where $val(i_1,\ldots,i_d) = \sum_{1 \leq j \leq d} 2^{j-1} i_j $, the value of the binary number given by the $i$'s.
Repeatedly applying the above argument, we get that the number of auxiliary variables that we need for an arbitrary coefficient is at most $m \cdot \log D = O(m \cdot \log(\text{deg}(f))) = \text{poly}(n)$.