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, by Chebyshev's inequality,
$$P(S\ne1)=P(|S-1|\ge1) \\ 
\le E(S-1)^2 \\ 
=ES^2-2ES+1\\ 
=\sum_{j,k\in[d]}EX_jX_k-2\sum_{j\in[d]}EX_j+1 \\ 
=d^2p^2+dpq-2dp+1\\ 
=1-1/d,\tag{1}$$
whence 
$$P(S=1)\ge1/d,\tag{2}$$
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. Indeed, consider the events 
$$B_j:=\{X_j=1,S-X_j=0\}=\{X_j=1,S=1\},\\
C_{j,k}:=\{X_j=X_k=1,S-X_j-X_k=0\}=\{X_j=X_k=1,S=2\}$$
for $j,k$ in $[d]$ such that $j<k$. These events are mutually exclusive, their union is the event $\{S\in\{1,2\}\}$, and the number of these events (that is, of the $B_j$'s and the $C_{j,k}$'s) is $n_d:=d+d(d-1)/2\le d^2$. Therefore, we may assign probabilities to these events as follows: 
$$P(B_j):=1/d^2,\quad P(C_{j,k}):=1/d^2,$$
so that $P(S=1)+P(S=2)=n_d/d^2\le1$, with $P(S=0):=1-n_d/d^2$. Then 
$P(S\in\{0,1,2\})=1$, so that the inequality in (1) becomes the equality, and hence the inequality in (2) becomes the equality.