There are the following formula in this link(https://en.wikipedia.org/wiki/Ramanujan%27s_congruences).
$$\sum_{k=0}^{ \infty } p(5k+4)q^k = 5 \frac{(q^5)_{\infty}^{5}}{(q)_{\infty}^{6}}.$$
$$\sum_{k=0}^{ \infty } p(7k+5)q^k = 7 \frac{(q^7)_{\infty}^{3}}{(q)_{\infty}^{4}} + 49q \frac{(q^7)_{\infty}^{7}}{(q)_{\infty}^{8}}.$$
Srinivasa Ramanujan discovered the following congruences.
$$ p(11k+6) = 0 mod.11.$$
Do you know the formla of $\sum_{k=0}^{ \infty } p(11k+6)q^k $?
That is
$$\sum_{k=0}^{ \infty } p(11k+6)q^k = ???.$$
Yes, a formula for the series $\sum_{k=0}^{ \infty } p(11k+6)q^k $, written as $\sum_{n=1}^{ \infty } p(11n-5)q^n$, can be found in the paper Ramanujan's unpublished manuscript on the partition and tau functions with proofs and commentary by Bruce C. Berndt and Ken Ono, on page $39$, formula $(18.7)$:
$$ \sum_{n=1}^{ \infty } p(11n-5)q^n(q_{11};q_{11})_{\infty}=11\sigma_7(n)q^n+121J, $$ which can be rewritten as you want, I suppose.