Principal specialisation on the forgotten symmetric polynomial looks exceptional, why?

Question

for any symmetric polynomial $u(\lambda,n)$ the principal specialisation is usually defined as $ps(u(x_1,x_2,x_3, ..)) = u(1,q,q^2,q^3, ..)$. See also https://www.math.upenn.edu/~peal/polynomials/standardSymmetricFunctions.htm.
I was surprised to find (conjecture) that the forgotten symmetic polynomials are exceptional in that $[n]_q! (1-q)^n \sum_{\lambda \vdash n} ps( f(\lambda) ) = q^{n(n-1)/2}$
All other symmetric functions produce less simple expressions. Why does the forgotten s.f. stand out in this regard?

Answer 1

I can add a few related observations (conjectures): $$ \sum_\lambda ps(m_\lambda) = \frac{1}{[n]_q!(1-q)^n} $$

A partition is even, if all parts are even.

For even $n$, $$ \sum_{\lambda: even } ps(f_\lambda) = \frac{ q^{(n/2)(n/2-1)} }{[2]_q [4]_q \dotsm [n]_q (q-1)^{n/2}} $$ and the sum is $0$ otherwise.

For even $n$, $$ \sum_{\lambda: even } ps(m_\lambda) = \frac{1}{[2]_q [4]_q \dotsm [n]_q (1-q)^{n/2}} \text{ if $n$ even} $$ and the sum is $0$ otherwise.