$\DeclareMathOperator\erf{erf}$
Following on from Milten's answer, we can consider expressions like these over a sequence of integers $(a_n)$. In particular we have a formula for the means $\mu_1,\mu_2,\dots$, where $\mu_n$ is defined as the arithmetic mean of all the expressions of the form $a_1\ast a_2\ast\cdots\ast a_n$ where each $\ast$ stands for $+$ or $\times$, namely
$$\mu_n=\frac{h(n)}{2^{n-1}}+\sum_{k=1}^{n-1}\frac{h(k)}{2^k}$$
where
$$h(n)=\prod_{k=1}^na_k+\frac{1}{2}\sum_{k=1}^{n-1}2^k\left(\prod_{r=k+1}^na_r\right).$$
For $(a_n)=(n)$, this formula gives us an asymptotic formula for $\mu_n$. We have
$$\mu_n\sim(e^2+1)\frac{n!}{2^n}.$$
To show this we will consider the sequence $(S_n)$ given by the sum of all the expressions $a_1\ast a_2\ast\cdots\ast a_n$, e.g. $S_1=a_1$, $S_2=(a_1+a_2)+(a_1\times a_2)$, etc. Note $\mu_n=S_n/2^{n-1}$.
We can find a recursive formula for $S_n$ by considering the possible forms of $a_1\ast a_2\ast\cdots\ast a_n$. In particular, we are summing expressions of the form:
\begin{align*}
1~\colon~&a_1\ast a_2\ast\cdots\ast a_{n-1}+a_n\\
2~\colon~&a_1\ast a_2\ast\cdots\ast a_{n-2}+a_{n-1}\times a_n\\
3~\colon~&a_1\ast a_2\ast\cdots\ast a_{n-3}+a_{n-2}\times a_{n-1}\times a_n\\
\vdots\\
n-1~:~&a_1+a_2\times\cdots\times a_n\\
n~:~&a_1\times\cdots\times a_n
\end{align*}
The sum of all the expressions of form $1$ is $S_{n-1}+2^{n-2}a_n$ because there are $2^{n-2}$ expressions of form $1$, likewise the sum of all the expressions of form $2$ is $S_{n-2}+2^{n-3}a_{n-1}a_n$, and so on until the expressions of form $n$ whose sum is $a_1a_2\cdots a_n$. It follows that
$$S_n=\prod_{k=1}^na_k+\sum_{k=1}^{n-1}\left(S_{n-k}+2^{n-1-k}\prod_{r=0}^{k-1}a_{n-r}\right).$$
Reversing the order of summation and re-indexing the product inside the sum we get
\begin{align*}
S_n&=\prod_{k=1}^na_k+\sum_{k=1}^{n-1}\left(S_k+2^{k-1}\prod_{r=k+1}^na_r\right)\\
&=\prod_{k=1}^na_k+\frac{1}{2}\sum_{k=1}^{n-1}\left(2^k\prod_{r=k+1}^na_r\right)+\sum_{k=1}^{n-1}S_k\\
&=h(n)+\sum_{k=1}^{n-1}S_k.
\end{align*}
Where $h(n)$ is defined as above. Repeated applications of this recurrence yield a formula for $S_n$,
$$S_n=h(n)+2^{n-1}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}$$
and so
$$\mu_n=\frac{h(n)}{2^{n-1}}+\sum_{k=1}^{n-1}\frac{h(k)}{2^k}.$$
If $(a_n)$ is such that $a_n\geq n$ for all $n\geq 1$, we can obtain an asymptotic formula for $\mu_n$, in particular I claim that
$$\mu_n\sim\frac{h(n)}{2^{n-1}}.$$
To show this, we must show that
$$\lim_{n\rightarrow\infty}\left(\frac{2^{n-1}}{h(n)}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}\right)=0.$$
We have
\begin{align*}
h(n)&=\prod_{k=1}^na_k+\frac{1}{2}\sum_{k=1}^{n-1}2^k\left(\prod_{r=k+1}^na_r\right)\\
&=a_n\left(\prod_{k=1}^{n-1}a_k+\frac{1}{2}\sum_{k=1}^{n-1}2^k\left(\prod_{r=k+1}^{n-1}a_r\right)\right)\\
&=a_n\left(\prod_{k=1}^{n-1}a_k+\frac{1}{2}\sum_{k=1}^{n-2}2^k\left(\prod_{r=k+1}^{n-1}a_r\right)+2^{n-2}\right)\\
&=a_n\left(h(n-1)+2^{n-2}\right)\\
&\geq a_nh(n-1).
\end{align*}
We have $a_n\geq n$ for all $n\geq1$ and hence
\begin{align*}
h(n)&\geq nh(n-1)\\
&\geq n(n-1)h(n-2)\\
&~~\vdots\\
&\geq n(n-1)\cdots(k+1)h(k)\\
&=\frac{n!}{k!}h(k).
\end{align*}
From which it follows
$$\frac{h(k)}{h(n)}\leq\frac{k!}{n!}$$
and so for all $n\geq2$,
\begin{align*}
\frac{2^{n-1}}{h(n)}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}&\leq \frac{2^{n-1}}{n!}\sum_{k=1}^{n-1}\frac{k!}{2^k}\\
&=\frac{2}{n}\left(\frac{1}{2}+\frac{2^{n-2}}{(n-1)!}\sum_{k=1}^{n-2}\frac{k!}{2^k}\right)\\
&\leq \frac{2}{n}\left(\frac{1}{2}+\frac{2^{n-2}}{(n-1)!}\sum_{k=1}^{n-2}\frac{(n-2)!}{2^{n-2}}\right)\\
&=\frac{2}{n}\left(\frac{1}{2}+\frac{n-2}{n-1}\right)\\
&\leq\frac{2}{n}\left(\frac{1}{2}+1\right)\\
&=\frac{3}{n}.
\end{align*}
The third line is justified by the fact that $k!/2^k$ is monotonically increasing.
Hence $$\lim_{n\rightarrow\infty}\left(\frac{2^{n-1}}{h(n)}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}\right)=0$$
and
$$\mu_n\sim\frac{h(n)}{2^{n-1}}.$$
Since $(a_n)$ is a sequence of positive integers, we can write
$$h(n)=\left(1+\frac{1}{2}\sum_{k=1}^{n-1}\frac{2^k}{\prod_{r=1}^ka_r}\right)\prod_{k=1}^na_k.$$
The sum in this expression for $h(n)$ converges because every term is positive and
$$\sum_{k=1}^{n-1}\frac{2^k}{\prod_{r=1}^ka_r}\leq\sum_{k=1}^{n-1}\frac{2^k}{k!}\leq e^2-1.$$
Thus
$$h(n)\sim\left(1+\frac{1}{2}\sum_{k=1}^{\infty}\frac{2^k}{\prod_{r=1}^ka_r}\right)\prod_{k=1}^na_k$$
and
$$\mu_n\sim\left(2+\sum_{k=1}^{\infty}\frac{2^k}{\prod_{r=1}^ka_r}\right)\frac{1}{2^n}\prod_{k=1}^na_k.$$
The following table shows the asymptotic behaviour of $\mu_n$ for various $(a_n)$.
$$\begin{array}{|c|c|}\hline
a_n & \sim\mu_n \\ \hline
n & \displaystyle{(e^2+1)\frac{n!}{2^n}} \\ \hline
mn & \displaystyle{(e^{2/m}+1)\left(\frac{m}{2}\right)^nn!} \\ \hline
2n-1 & \displaystyle{\left(1+e\sqrt{\pi}\erf(1)\right)\frac{(2n)!}{2^{2n}n!}} \\\hline
2^n & \displaystyle{\approx(3.64163)2^\frac{n(n-1)}{2}} \\\hline
2^{n^2} & \displaystyle{\approx(3.12549)2^\frac{(2n+5)n(n-1)}{6}} \\\hline
n^2 & \displaystyle{\approx(5.25235)\frac{n!^2}{2^n}} \\\hline
n^{11} & \displaystyle{\approx(4.00195)\frac{n!^{11}}{2^n}} \\\hline
F_n & \displaystyle{\approx(19.6833)\frac{\varphi^{\frac{n(n+1)}{2}}}{(2\sqrt{5})^n}} \\\hline
1 & \displaystyle{\frac{n+1}{2}\quad(\textrm{Exact})} \\\hline
2 & \displaystyle{\frac{n^2+5n+2}{4}\quad(\textrm{Exact})} \\\hline
\end{array}$$
Note $F_n$ is the $n$th Fibonacci number and $\varphi$ is the golden ratio. Since $F_2< 2$, we cannot immediately use the asymptotic formula above, the trick is to use the law of total expectation to find that $\mu_n=\mu_{n-1}^*+1/2$, where $\mu_n^*$ is the sequence of means associated to $(F_{n+1})$. In the end we find that the asymptotic formula works just the same for $(F_n)$, which shows that the condition given for the asymptotic formula to hold is not a necessary one.
Remarks
I suspect a similar method to this could be used to determine the behaviour of the variance, at least for some choices of $(a_n)$. I have no idea about determining which values are taken more than once by such expressions, I suppose one might want to consider the primitive cases of this, i.e. $1+2+3=1\times2\times3$ is primitive while $1+2+3+4=1\times2\times3+4$ is not.
Edit
There is a much stronger result for the asymptotic behaviour of such sequences, namely:
If $(a_n)$ is a sequence of non-negative real numbers such that
$$\lim_{n\rightarrow\infty}a_n=\infty,$$
then
$$\mu_n\sim h(n)/2^{n-1}.$$
Proof
For all $M\in\mathbb{R}$, there is an $N\in\mathbb{N}$ such that for all $n\geq N$, $a_n>M$. Let $M>2$, then using the recurrence for $h(n)$ from before, we have for all $n> k\geq N$
\begin{align*}
h(n)&=a_n(h(n-1)+2^{n-2})\\
&\geq Mh(n-1)\\
&~~\vdots\\
&\geq M^{n-k}h(k).
\end{align*}
If $h(n)=0$ then $h(n+1)=a_{n+1}2^{n-1}>0$, so $h(n)\neq0$ for all sufficiently large $n$, thus
$$\frac{h(k)}{h(n)}\leq M^{k-n}.$$
Again for $n$ sufficiently large
\begin{align*}
\frac{2^{n-1}}{h(n)}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}&\leq\frac{2^{n-1}}{h(n)}\left(\sum_{k=1}^N\frac{h(k)}{2^k}+\sum_{k=N+1}^{n-1}\frac{h(k)}{2^k}\right)\\
&\leq\frac{2^{n-1}}{h(n)}\sum_{k=1}^N\frac{h(k)}{2^k}+2^{n-1}\sum_{k=N+1}^{n-1}\left(\frac{M^{k-n}}{2^k}\right)\\
&\leq \frac{2^{n-1}}{h(n)}\sum_{k=1}^N\frac{h(k)}{2^k}+\frac{1}{2}\left(\frac{2}{M}\right)^n\sum_{k=0}^{n-1}\left(\frac{M}{2}\right)^k\\
&=\frac{2^{n-1}}{h(n)}\sum_{k=1}^N\frac{h(k)}{2^k}+\frac{1}{2}\left(\frac{2}{M}\right)^n\frac{\left(\frac{M}{2}\right)^n-1}{\frac{M}{2}-1}\\
&= \frac{2^{n-1}}{h(n)}\sum_{k=1}^N\frac{h(k)}{2^k}+\frac{1-\left(\frac{2}{M}\right)^n}{M-2}\\
&\leq \frac{2^{n-1}}{h(n)}\sum_{k=1}^N\frac{h(k)}{2^k}+\frac{1}{M-2}.
\end{align*}
The recurrence for $h(n)$ tells us that $2^{n-1}/h(n)\leq2/a_n$, so
$$\frac{2^{n-1}}{h(n)}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}\leq\frac{2}{a_n}\sum_{k=1}^N\frac{h(k)}{2^k}+\frac{1}{M-2}.$$
For sufficiently large $n$,
$$\frac{2}{a_n}\sum_{k=1}^N\frac{h(k)}{2^k}\leq\frac{1}{M},$$
and so
$$\frac{2^{n-1}}{h(n)}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}\leq\frac{1}{M}+\frac{1}{M-2}.$$
Finally, letting $M\rightarrow\infty$, and hence also $n\rightarrow\infty$, we have the result
$$\lim_{n\rightarrow\infty}\left(\frac{2^{n-1}}{h(n)}\sum_{k=1}^{n-1}\frac{h(k)}{2^k}\right)=0.$$
If $a_n\neq0$ for all $n\geq1$, we may write
$$\mu_n\sim\left(2+\sum_{k=1}^{\infty}\frac{2^k}{\prod_{r=1}^ka_r}\right)\frac{1}{2^n}\prod_{k=1}^na_k.$$
Note the sum converges because from some point on $a_n>2$.
The same result probably also holds if $a_n$ can take negative values, but the proof seems a bit finicky and I had limited time.
