Not sure if this helps, but the determinantal expression for $\text{sym}^n_t f$ is in fact a quantum determinant --- albeit of a scalar-valued matrix and not a quantum matrix (whose entries must satisfy the Faddeev-Reshetikhin-Takhtadzhyan relations). For example, when $n=4$ we have:

\begin{equation} \det \, \begin{pmatrix} {1 \over 4}f_1 & {1 \over 4}f_2 & {1 \over 4}f_3 & {1 \over 4}f_4 \\ -t & {1 \over 3}f_1 & {1 \over 3}f_2 & {1 \over 3}f_3 \\ 0 & -t & {1 \over 2}f_1 & {1 \over 2}f_2 \\ 0 & 0 & -t & f_1\end{pmatrix} \ = \ \text{det}_q \, \begin{pmatrix} {1 \over 4}f_1 & {1 \over 4}f_2 & {1 \over 4}f_3 & {1 \over 4}f_4 \\ -1 & {1 \over 3}f_1 & {1 \over 3}f_2 & {1 \over 3}f_3 \\ 0 & -1 & {1 \over 2}f_1 & {1 \over 2}f_2 \\ 0 & 0 & -1 & f_1\end{pmatrix} \end{equation}

yours, Ines

Not sure if this helps, but the determinantal expression for $\text{sym}^n_t f$ is in fact a quantum determinant --- albeit of a scalar-valued matrix and not a quantum matrix (whose entries must satisfy the Faddeev-Reshetikhin-Takhtadzhyan relations). For example, when $n=4$ we have:

\begin{equation} \det \, \begin{pmatrix} {1 \over 4}f_1 & {1 \over 4}f_2 & {1 \over 4}f_3 & {1 \over 4}f_4 \\ -t & {1 \over 3}f_1 & {1 \over 3}f_2 & {1 \over 3}f_3 \\ 0 & -t & {1 \over 2}f_1 & {1 \over 2}f_2 \\ 0 & 0 & -t & f_1\end{pmatrix} \ = \ \text{det}_t \, \begin{pmatrix} {1 \over 4}f_1 & {1 \over 4}f_2 & {1 \over 4}f_3 & {1 \over 4}f_4 \\ -1 & {1 \over 3}f_1 & {1 \over 3}f_2 & {1 \over 3}f_3 \\ 0 & -1 & {1 \over 2}f_1 & {1 \over 2}f_2 \\ 0 & 0 & -1 & f_1\end{pmatrix} \end{equation}

yours, Ines