$(f \star f \star f) (x) $ at $x = 0$ $$g(x) = (f \star f ) (x)$$ and $$q(x) =  (f \star f  \star f) (x)$$ where $\star$ denotes convolution.
I can write $$g(x) = \int\limits_{-\infty}^{\infty}f(\tau)f(\tau - x) d \tau$$
Therefore $$g(0) = \int\limits_{-\infty}^{\infty}f(x)f(x) dx$$
Is there a similar simplification for $q(0)$?
I started with $$q(0) = \int\limits_{-\infty}^{\infty}f(\lambda) \left(\int\limits_{-\infty}^{\infty} f(\tau)f(\tau - \lambda) d \tau \right) d \lambda$$
Can this be further simplified? Is there a possibility of expressing $q(0)$ with an integral involving one variable (like $g(0)$)?
 A: Firstly, your formulas for $g(x)$ and $q(x)$ are incorrect, which are giving you the wrong impression. It's crucial for convolutions that $\tau$ goes up and $-\tau$ goes down as the integration variable increases. We have: $g(x)=\displaystyle{\int_{-\infty}^{\infty}}f(\tau)f(x-\tau)\,\mathrm{d}\tau$ and similarly for $q$.  Therefore, the tidy formula for $g(0)$ is actually $\displaystyle{\int_{-\infty}^{\infty}}f(x)f(-x)\,\mathrm{d}x$, which only equals your expression if $f$ is even.

For $(f\star f\star f)(0)$ you would take a formula for triple convolution (just imagine that $\mathbb{T}$ at that link is replaced with $(-\infty,\infty)$) and substitute in $0$.
We get formulas like $\displaystyle{\int_{-\infty}^{\infty}}f(y)\displaystyle{\int_{-\infty}^{\infty}}f(x)f(-x-y)\,\mathrm{d}x\,\mathrm{d}y$ and $\displaystyle{\int_{-\infty}^{\infty}}f(-y)\displaystyle{\int_{-\infty}^{\infty}}f(x)f(y-x)\,\mathrm{d}x\,\mathrm{d}y$
You can move the outer $f(y)$ or $f(-y)$ inside the inner integral if you like, but unless $f$ is even or has some other special property, you're not going to get something simpler.
