4 added 162 characters in body

It would help to know what $L$ is.

update This answer was for the question of how short the longest projection could be if $L$ is allowed to vary. There have since been some clarifications.

Here is an upper bound which I think I can prove to be the minimum among all the $d$ dimensional spaces of $\mathbb{R}^n$ (with $n>d$ of course)

Let $L$ be the subspace of $\mathbb{R}^n$ consisting of vectors whose first $d+1$ coordinates sum to $0$ and whose remaining coordinates are $0$. Project the vectors of $\lbrace0,1\rbrace^n$ onto $L.$ The length of the longest projection is $$\begin{cases} \sqrt{\frac{d+1}{4}}, & \mbox{if }d=2k+1 \\ \\ \sqrt{\frac{d+1-\frac{1}{d+1}}{4}}, & \mbox{if }d=2k. \end{cases}$$

3 added 14 characters in body

It would help to know what $L$ is.

Here is an upper bound which I think I can prove to be the minimum among all the $d$ dimensional spaces of $\mathbb{R}^n$ (with $n>d$ of course)

Let $L$ be the subspace of $\mathbb{R}^n$ consisting of vectors whose first $d+1$ coordinates sum to $0$ and whose remaining coordinates are $0$. Project the vectors of $\lbrace0,1\rbrace^n$ onto $L.$ The length of the longest projection is $$\begin{cases} \sqrt{\frac{d}{4}}, sqrt{\frac{d+1}{4}}, & \mbox{if }d=2k d=2k+1 \\ \\ \sqrt{\frac{d-1/d}{4}}, sqrt{\frac{d+1-\frac{1}{d+1}}{4}}, & \mbox{if }d=2k+1. d=2k. \end{cases}$$

2 added 14 characters in body

It would help to know what $L$ is.

Here is an upper bound which I think I can prove to be the minimum among all the $d$ dimensional spaces of $\mathbb{R}^n$ (with $n>d$ of course)

Let $L$ be the subspace of $\mathbb{R}^n$ consisting of vectors whose first $d+1$ coordinates sum to $0$ and whose remaining coordinates are $0$. Project the vectors of $\lbrace0,1\rbrace^n$ onto $L.$ The length of the longest projection is $$\begin{cases} \frac{d}{4}, sqrt{\frac{d}{4}}, & \mbox{if }d=2k \\ \\ \frac{d-1/d}{4}, sqrt{\frac{d-1/d}{4}}, & \mbox{if }d=2k+1. \end{cases}$$

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