This is not an answer, but just somewhere useful you could start looking. Its good to think about when the quotient of fundamental groups has more than one generator.

The Rayleigh principle should imply that $\lambda_1(M,g)\geq \lambda_1(\tilde{M},\tilde{g})$, since your $\lambda_1(M,g)=\inf_{f\in H^1(M), ||f||_{2,g}=1}\int_M|df|^2$, and this integral is multiplicative under finite covers. (So this infimum upstairs can only be smaller, as there may be $H^1(M)$ functions which weren't lifts by $p$).

In this paper, https://projecteuclid.org/euclid.tmj/1178224610, they prove that when the covering $p:\tilde{M}\rightarrow M$ satisfies $\pi_1(M)/p_\ast\pi_1(\tilde{M})\simeq \mathbb{Z}_k$, there exists a metric on $M$ such that $\lambda_1(M,g)=\lambda_1(\tilde{M},\tilde{g})$. But this may not hold in general for finite covers.

This is not an answer, but just somewhere useful you could start looking. Its good to think about when the quotient of fundamental groups has more than one generator.

The Rayleigh principle should imply that $\lambda_1(M,g)\geq \lambda_1(\tilde{M},\tilde{g})$, since your $\lambda_1(M,g)=\inf_{f\in H^1(M), ||f||_{2,g}=1}\int_M|df|^2$, and this integral is multiplicative under finite covers. (So this infimum upstairs can only be smaller, as there may be $H^1(M)$ functions which weren't lifts by $p$).

In this paper, https://projecteuclid.org/euclid.tmj/1178224610, they prove that when the covering $p:\tilde{M}\rightarrow M$ satisfies $\pi_1(M)/p_\ast\pi_1(\tilde{M})\simeq \mathbb{Z}_k$, there exists a metric on $M$ such that $\lambda_1(M,g)=\lambda_1(\tilde{M},\tilde{g})$. But this may not hold general finite covers.