The highest lower bound is $1/d$. 

Indeed, for each $j\in[d]:=\{1,\dots,d\}$, let  $A_j$ denote the event of the head on the $j$th coin and let $X_j:=1_{A_j}$. Let $S:=X_1+\dots+X_d$. Then the event of getting exactly one head is $P(S=1)$. 

Note that $EX_j=p$ and $EX_jX_k=p^2+pq\,1(j=k)$ for all $j,k$ in $[d]$, where $p:=1/d$ and $q:=1-p$. So,
$$P(S\ne1)=P(|S-1|\ge1)\le E(S-1)^2=ES^2-2ES+1=\sum_{j,k\in[d]}EX_jX_k-2\sum_{j\in[d]}EX_j+1=d^2p^2+dpq-2dp+1=1-1/d,$$
whence 
$$P(S=1)\ge1/d,$$
so that $1/d$ is indeed a lower bound on the probability of getting exactly one head. 

It is also easy to see that this lower bound is attained.