Not a closed form, but we have the following series representations:
$$\begin{align*} p(x) = \prod_{k=0}^{\infty} \left( 1-x^{2^k} \right)
&= \frac1{1-x}\left(1+2\sum_{k\geq0}\frac{(-1)^kx^{2^k}}{1+x^{2^k}}\right) \\
&= \frac{1}{1-x} - \frac{4x}{(1-x)^2} + \frac6{1-x} \sum_{k=0}^\infty \frac{x^{2^{2k}}}{1-x^{2^{2k+1}}}
\end{align*}$$
where the last sum is $$\sum_{\nu_2(n)\text{ even}}x^n$$
Proof. We have: (on the level of formal series; although everything converges absolutely for $|x|<1$)
$$p(x)-1 = \sum_{n=1}^\infty x^n (-1)^{b(n)}$$
where $b(n)$ is the number of $1$'s in the binary expansion of $n$. We can write:
$$(-1)^{b(n)} = 1+2\sum_{\substack{k \text{ for which the }\\k\text{th digit is 1,}\\\text{starting from }k=0}}(-1)^k$$
plug this in the sum and change the order of summation $n \leftrightarrow k$:
$$\begin{align*}p(x)-1
&= \sum_{n=1}^\infty x^n + 2 \sum_{n=1}^\infty x^n\sum_{\substack{k \text{ for which the }\\k\text{th digit in }n\text{ is 1}}}(-1)^k \\
&= \frac x{1-x} + 2 \sum_{k=0}^\infty (-1)^k x^{2^k} \prod_{j \neq k}(1+x^{2^j}) \\
&= \frac x{1-x} + 2 \sum_{k=0}^\infty (-1)^k \frac{x^{2^k}}{(1-x)(1+x^{2^k})} \\
\end{align*}$$
For the series $S = \sum_{k=0}^\infty (-1)^k \frac{x^{2^k}}{1+x^{2^k}}$, we can write each term as a geometric series:
$$S = -\sum_{k=0}^\infty \sum_{n=1}^\infty (-1)^{k+n}x^{2^kn}$$
Grouping the terms with the same power of $x$ gives:
$$\sum_{n=1}^\infty c_nx^n$$
where $c_n=1$ if $\nu_2(n)$ is even, and $c_n=-2$ if $\nu_2(n)$ is odd.
We get $$\begin{align*}S
&= -2\frac x{1-x} + 3\sum_{\nu_2(n)\text{ even}}x^n \\
&= -2\frac x{1-x} + 3\sum_{k=0}^\infty \left( (x^{4^k})^1 + (x^{4^k})^3 + (x^{4^k})^5 + \cdots \right)\\
&= -2\frac x{1-x} + 3\sum_{k=0}^\infty \frac{x^{2^{2k}}}{1-x^{2^{2k+1}}}
\end{align*}$$
