About the group of units of $F_2G$ Let $F_2G$ be the group algebra generated by group elements of $G$ over the field of characteristic $2$. Refer to it as the group ring $RG$ where $R=F_2$ and $G$ is any commutative or non-commutative group.
It is well known that every element $a\in RG$ has an embedding into $M_{n\times n}(R)$ the ring of $n \times n$ matrices over $R$. Then we can identify it with the following morphism:
$$\phi : RG \mapsto M_{n\times n}(R)$$
Then $a \in RG$ is an unit $\iff Det(\phi(a))=Det(M_a) \in U(R)$ this is only if the determinant of $M_a$ is invertible in $R$.
With this information we can easiliy test if an element of $F_2G$ is an unit, but what's the order in general when $R=F_2$ and $G$ non-commutative?
If I randomly construct a list of $n$ elements in $F_2G$ what's the probability that an element of this list is on $U(F_2G)$?
 A: This is an answer to your last question in representation-theoretic terms for a special case. As already mentioned in the comments, the general case seems to be quite hard.  
Suppose that $G$ is a finite group with a normal 2-Sylow subgroup $H$. Then the Jacobson radical of $F_2[G]$ is the kernel of $F_2[G] \to F_2[G/H]$. Now $F_2[G/H]$ is semisimple and we have an isomorphism $F_2[G/H] \cong \prod_{i}^r M_{n_i}(\Bbb F_{2^{k_i}})$ where $r$ is the number of irreducible representations. In general, if $R$ is a ring, then any unit of $R/J(R)$ lifts to a unit of $R$: suppose $x,y \in R$, such that $xy,yx \in 1+J(R) \subset R^\times$ then we find $u,v \in R$ with $xyu=1, vyx=1$, showing that $x$ is left and right invertible. This implies that we have a short exact sequence:
$$1 \to 1 + J(R) \to R^\times \to R/J(R)^\times \to 1$$ so that if $R$ is finite, we have $|R^\times|=|J(R)||R/J(R)^\times|$.
In our setting, we have a surjective map $F_2[G] \to F_2[G/H]$ with kernel $J(F_2[G])$, so that $|J(F_2[G])|=2^{|G|}/2^{|G/H|}=2^{|G|-|G/H|}$.
Now because $F_2[G/H]$ is semisimple, we can characterize $|F_2[G/H]^\times|$ in terms of the representation theory. If we let $V_1, \dots, V_r$ be the irreducible representation of $G$ (or equivalently, of $G/H$) over $F_2$ and let $\mathrm{dim}_{F_2}V_i = n_i$ and $\mathrm{dim}_{F_2}\mathrm{End}_{F_2[G]}(V_i)=k_i$, then we have the Artin-Wedderburn decomposition $$F_2[G/H] \cong \prod^{r}_i M_{n_i}(F_{2^{k_i}})$$ implying that $$F_2[G/H]^\times \cong \prod_{i=1}^r \mathrm{GL}_{n_i}(2^{k_i})$$
Now as $$|\mathrm{GL}_{n_i}(2^{k_i})|=\prod_{j=0}^{n_i-1}(2^{n_i k_i}-2^{k_i j})$$ we obtain $$|F_2[G]^\times|=2^{|G|-|G|/|H|} \prod_{i=1}^r \prod_{j=0}^{n_i-1}(2^{n_i k_i}-2^{k_i j})$$ thus the probability $p$, that a randomly chosen element from $F_2[G]$ is a unit is that divided by $2^{|G|}$, i.e. $$p=2^{-|G|/|H|} \prod_{i=1}^r \prod_{j=0}^{n_i-1}(2^{n_i k_i}-2^{k_i j})$$
By a standard probability argument, the probability that from a list of $n$ elements from $F_2[G]$, at least one is a unit is $1-(1-p)^n$
I'm not sure if this is the kind of answer you're looking for and it is only for a special case, but it's something I guess.
