I will assume that you are considering symmetric as part of the definition of positive semi-definite; otherwise there are easy counterexamples like $$A=\begin{bmatrix}1&2\\0&1\end{bmatrix}.$$
When $A$ is positive semi-definite and symmetric, consider first the $2\times2$ case. We can write $A=Q^TDQ$, where $Q$ is orthogonal and $D$ is diagonal with diagonal given by the eigenvalues $\lambda,\mu$ of $A$. The positivity of $A$ together with $\|A\|≤1$ give us $0≤\lambda,\mu≤1$. Writing $$Q=\begin{bmatrix}\cos t&\sin t\\-\sin t&\cos t\end{bmatrix}$$ for an appropriate $t$ we get, after calculating the entries of $Q^TDQ$,
$$
a_{11}+a_{12}+a_{21}+a_{22}=\lambda+\mu+(\lambda-\mu)\cos t\sin t.
$$
Assuming without loss of generality that $\lambda≥\mu$, the crude estimate $-1≤\cos t\sin t≤1$ gives us
$$
2\mu\leq a_{11}+a_{12}+a_{21}+a_{22}≤2\lambda.
$$
Thus
$$\tag1
a_{11}+a_{12}+a_{21}+a_{22}\leq2.
$$
When $A$ is $n\times n$, we can write$\def\tr{\operatorname{Tr}}$
$$
\tr(A_1+A_2+A_3+A_4)=\sum_{k=1}^{n/2}(A_1)_{kk}+(A_2)_{kk}+(A_3)_{kk}(A_4)_{kk}.
$$
The $2\times2$ matrix $$\begin{bmatrix}(A_1)_{kk}&(A_2)_{kk}\\ (A_3)_{kk}& (A_4)_{kk}\end{bmatrix}$$ is a principal submatrix of $A$ and hence positive semi-definite (and it's norm is at most $1$, seen either by Cauchy Interlacing or by noticing that the submatrix is of the form $PAP$ with $P$ an orthogonal projection and hence $\|PAP\|≤\|A\|$). Then, using $(1)$,
$$\tag2
\tr(A_1+A_2+A_3+A_4)\leq\sum_{k=1}^{n/2}2=n.
$$
As for the diagonal dominance, it is not true in general. What can be said is the following: when $A$ is positive semi-definite and symmetric, there exists $X$, such $A=X^TX$. This gives us, using Cauchy-Schwarz, the inequality $$|a_{kj}|≤a_{kk}^{1/2}a_{jj}^{1/2},$$
which comes from
$$
|a_{kj}|=\Big|\sum_gx_{gk}x_{gj}\Big|≤\Big(\sum_gx_{gk}^2\Big)^{1/2} \Big(\sum_gx_{gj}^2\Big)^{1/2}= a_{kk}^{1/2}a_{jj}^{1/2}.
$$
Since $$a_{kk}^{1/2}a_{jj}^{1/2}\leq\frac{a_{kk}+a_{jj}}2,$$ we get that
$$
\tr(A_2)\leq\frac{\tr(A_1+A_4)}2.
$$
The inequality doesn't look to be sharp, though, as the maximum in $(2)$ seems to be achieved when $A$ is diagonal. For example, when $a_{kj}=\frac1n$, we have
$$
\tr(A_1+A_2+A_3+A_4)=\sum_{k=1}^{n/2}\frac4n=2.
$$
