Yes, it is the same as $\mathbb{F}$.

As John Baez points out, it is the same as the free symmetric monoidal $\infty$-groupoid on one object. (This can also be seen by playing around with the adjoints between groupoids and categories).

Symmetric monoidal $\infty$-groupoids are the same as $E_\infty$-spaces (careful: not assumed to be "grouplike"). So we want the free $E_\infty$-space on a point. As a space this is homotopy equivalent to $\sqcup_{n\geq0} B \Sigma_n$. Note all the hom spaces are actually 1-types, so it can be modeled as a 1-groupoid, namely $\mathbb{F}$. It might seem surprising at first that the free symmetric monoidal $\infty$-category is actually an ordinary category. However on reflection we see that it follows from the fact that the spaces in the $E_\infty$-operad are actually 1-types.

Equivalently, as John says, it is the space of all finite subsets of $\mathbb{R}^\infty$. If you pick an identification of $\mathbb{R}^\infty$ with the "interior" of an infinite cube, then you get a natural $E_\infty$ structure on this model of the space, too.

You can also see that it is $\mathbb{F}$ in the way that Denis suggests in his comment to the OP.

commutatively monoidalobject. (If you take, say, $\mathbb F$ with cartesian product $\times$ and an empty set $\varnothing$ instead, then there is no monoidal functor from $(\mathbb F,\uplus)$ to $(\mathbb F,\times)$ that maps $\ast$ to $\varnothing$. The unique function from $\varnothing$ to $\ast$ would have to be mapped to a function from $\ast$ to $\varnothing$, and there is none.) $\endgroup$groupoidof finite sets, which of I'm not mistaken is the universal symmetric monoidal $\infty$-category on one object since it's the symmetric monoidal envelope of the $\infty$-operad $Triv^\otimes$ $\endgroup$