In view of the defining recursion, when $t=1$, the symmetric power $\text{sym}^n \psi$ has an elegant expression --- namely as the coefficient of $s^n$ in the exponent of "logarithmic" expansion
\begin{equation} \begin{array}{ll} &{\displaystyle \exp \Big( s \, \psi_1 \ + \ {s^2 \over 2} \, \psi_2 \ + \ {s^3 \over 3} \, \psi_3 \ + \ \cdots \Big)} \ &= \\ &{\displaystyle \exp \big( s \, \psi_1\big) \cdot \exp \big( {s^2 \over 2} \, \psi_2 \big) \cdot \exp \big({s^3 \over 3} \, \psi_3\big) \cdots } \ &= \end{array} \end{equation}
In the caes of the real field $\Bbb{R}$ and the additive character $\psi(x) = \exp (-x)$ we may expand the factor
\begin{equation} \exp \Big( {s^k \over k} \, \psi_k(x) \Big) \ = \sum_{r \geq 0} \ {s^{kr} \over {n! \, k^r}} \, \psi\big( {\scriptstyle -} \, rx^k \big) \end{equation}
If this expansion is formally recycled into the previous infinite product we may conclude that $\text{sym}^n \psi$ is the coefficient of $s^n$ in
\begin{equation} \sum_{ \text{p} } {s^{\, p_1} \cdot s^{\ p_2} \cdot s^{\, p_3} \cdots \, s^{p_d} \over { {\scriptstyle (p_1)! \,(p_2)! \,(p_3)! \, \cdots \, (p_d)!} \cdot {\scriptstyle 2^{p_2} \, 3^{p_3} \, \cdots \, d^{p_d} }}} \, \psi\big( {\scriptstyle -} \text{p} \big) \end{equation}
where the sum is taken over all polynomials $\text{p} = p_1x + p_2x^2 + \dots + p_dx^d$ with non-negative integer coefficients and with zero constant term. Alternatively we may encode such a polynomial $\text{p}$ as a composition ${\bf p} = \big(p_1, \dots, p_d\big) \,$ in which case the desired coefficient of $s^n$ is
\begin{equation} \sum_{ {\bf \text{p}} \, \vdash n } {2^{-p_2} \, \cdots \, d^{-p_d} \over { (p_1)! \, \cdots \, (p_d)!} } \, \psi \big( {\scriptstyle -} \text{p} \big) \end{equation}
where the sum is taken over all compositions ${\bf p} = \big(p_1, \dots, p_d\big) \,$ whose tally $p_1 + 2p_2 + 3p_3 + \dots + dp_d$ is exactly $n$.
yours, in crank-heit
A. Leverkühn